Black Box——优先队列
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Black Box
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1
N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.
Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.
Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.
Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6
Sample Output
3312
题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。
解析：
首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。
#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}
Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions: ADD (x): put element x into Black Box; GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending. Let us examine a possible sequence of 11 transactions: Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}Black BoxTime Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64uDescription
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box;
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.
Let us examine a possible sequence of 11 transactions:
Example 1N Transaction i Black Box contents after transaction Answer (elements are arranged by non-descending) 1 ADD(3) 0 3 2 GET 1 3 3 3 ADD(1) 1 1, 3 4 GET 2 1, 3 3 5 ADD(-4) 2 -4, 1, 3 6 ADD(2) 2 -4, 1, 2, 3 7 ADD(8) 2 -4, 1, 2, 3, 8 8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8 9 GET 3 -1000, -4, 1, 2, 3, 8 1 10 GET 4 -1000, -4, 1, 2, 3, 8 2 11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer arrays:
1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).
2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).
The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312题意：输入M个数字，在输入N个测试数据，第i个测试数据代表从头数a[i]个数后，这些数中的第i个小的数。解析：首先可以注意到，这N个测试数据都是递增的，所以就可以用一层循环去控制，在遍历M个数字所存在的数组的时候，就可以利用两个优先队列，一个优先队列cx以大优先，也就是队顶放整个队列中最大的元素，另一个队列cd以小优先，队顶放最小的元素。当输入第一个测试数据的时候，cx队列里没有东西，cd队列里的顶就是第一个最小的数，将其输出并放入cx队列里，释放top，在之后的测试数据中，将从M中拿出来的数据分别和cx与cd的顶比较，如果比cx的顶还小，则cx的顶放入cd中，并将这个元素放入cx中，如果比cx的顶大，则直接进入cd中，输出cd中的顶。#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#define MAX 100007using namespace std;struct cmp{ bool operator()(const int &a,const int& b) { return a > b;//队列的顶部放最小的元素，切记！！！！ }};int mapp[50000];int main(){ priority_queue<int> cx;//定义一个优先队列，以大优先，这也是默认的 priority_queue<int,vector<int>,cmp> cd;//以小优先，需要重载运算符 int n,m,i,j,k,a; while(scanf("%d%d",&n,&m)!=EOF) { for(i = 1; i <= n; i++) scanf("%d",&mapp[i]); while(!cx.empty()) cx.pop(); while(!cd.empty()) cd.pop(); i = 1; for(j = 1;j <= m;j++) { scanf("%d",&a); for(i;i<=a;i++) { if(j == 1) { cd.push(mapp[i]); } else { if(mapp[i] < cx.top()) { cd.push(cx.top()); cx.pop(); cx.push(mapp[i]); } else { cd.push(mapp[i]); } } } printf("%d\n",cd.top()); cx.push(cd.top()); cd.pop(); } } return 0;}
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