Codeforces Round #277.5 (Div. 2)部分题解

A. SwapSort

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

In this problem your goal is to sort an array consisting of n integers in at most n swaps. For the given array find the sequence of swaps that makes the array sorted in the non-descending order. Swaps are performed consecutively, one after another.

Note that in this problem you do not have to minimize the number of swaps — your task is to find any sequence that is no longer than n.

Input

The first line of the input contains integer n (1 ≤ n ≤ 3000) — the number of array elements. The second line contains elements of array: a0, a1, ..., an - 1 ( - 109 ≤ ai ≤ 109), where ai is the i-th element of the array. The elements are numerated from 0 to n - 1 from left to right. Some integers may appear in the array more than once.

Output

In the first line print k (0 ≤ k ≤ n) — the number of swaps. Next k lines must contain the descriptions of the kswaps, one per line. Each swap should be printed as a pair of integers i, j (0 ≤ i, j ≤ n - 1), representing the swap of elements ai and aj. You can print indices in the pairs in any order. The swaps are performed in the order they appear in the output, from the first to the last. It is allowed to print i = j and swap the same pair of elements multiple times.

If there are multiple answers, print any of them. It is guaranteed that at least one answer exists.

Sample test(s)

input

55 2 5 1 4

output

20 34 2

input

610 20 20 40 60 60

output

0

input

2101 100

output

10 1

排序之后贪心一下

<span style="font-size:18px;">/*************************************************************************    > File Name: cf.cpp    > Author: acvcla    > QQ:     > Mail: [e-mail]     > Created Time: 2014年11月17日 星期一 23时34分13秒 ************************************************************************/#include<iostream>#include<algorithm>#include<cstdio>#include<vector>#include<cstring>#include<map>#include<queue>#include<stack>#include<string>#include<cstdlib>#include<ctime>#include<set>#include<math.h>using namespace std;typedef long long LL;const int maxn = 3e3 + 10;#define rep(i,a,b) for(int i=(a);i<=(b);i++)#define pb push_backint A[maxn],B[maxn];std::vector<int>x,y;int main(){ios_base::sync_with_stdio(false);cin.tie(0);int n;int ans=0;while(cin>>n){x.clear();y.clear();for(int i=0;i<n;i++){cin>>A[i];B[i]=A[i];}sort(B,B+n);for(int i=0;i<n;i++){if(A[i]==B[i])continue;int M=0;for(int j=i+1;j<n;j++)if(A[j]==B[i]){M=j;if(B[j]==A[i]){swap(A[i],A[j]);x.pb(i);y.pb(j);break;}}if(A[i]==B[i])continue;swap(A[i],A[M]);x.pb(i);y.pb(M);}cout<<x.size()<<endl;for(int i=0;i<x.size();i++){cout<<x[i]<<' '<<y[i]<<endl;}}return 0;}</span>

B. BerSU Ball

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! n boys and m girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.

We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.

For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from n boys and m girls.

Input

The first line contains an integer n (1 ≤ n ≤ 100) — the number of boys. The second line contains sequencea1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the i-th boy's dancing skill.

Similarly, the third line contains an integer m (1 ≤ m ≤ 100) — the number of girls. The fourth line contains sequence b1, b2, ..., bm (1 ≤ bj ≤ 100), where bj is the j-th girl's dancing skill.

Output

Print a single number — the required maximum possible number of pairs.

Sample test(s)

input

41 4 6 255 1 5 7 9

output

3

input

41 2 3 4410 11 12 13

output

0

input

51 1 1 1 131 2 3

output

2

直接贪心就好

<span style="font-size:18px;">/*************************************************************************    > File Name: cf.cpp    > Author: acvcla    > QQ:     > Mail: [e-mail]     > Created Time: 2014年11月17日 星期一 23时34分13秒 ************************************************************************/#include<iostream>#include<algorithm>#include<cstdio>#include<vector>#include<cstring>#include<map>#include<queue>#include<stack>#include<string>#include<cstdlib>#include<ctime>#include<set>#include<math.h>using namespace std;typedef long long LL;const int maxn = 1e3 + 10;#define rep(i,a,b) for(int i=(a);i<=(b);i++)#define pb push_backint A[maxn],B[maxn];int main(){ios_base::sync_with_stdio(false);cin.tie(0);int n,m;while(cin>>n){memset(A,0,sizeof A);memset(B,0,sizeof B);int x;rep(i,1,n){cin>>x;++A[x];}cin>>m;rep(i,1,m){cin>>x;++B[x];}int ans=0;rep(i,1,100){if(A[i]<=0)continue;ans+=min(A[i],B[i-1]+B[i]+B[i+1]);if(A[i]>=B[i-1]+B[i]+B[i+1])B[i-1]=B[i]=B[i+1]=0;else{if(B[i-1]>0&&A[i]>0){int t=B[i-1];B[i-1]=max(B[i-1]-A[i],0);A[i]=max(A[i]-t,0);}if(B[i]>0&&A[i]>0){int t=B[i];B[i]=max(B[i]-A[i],0);A[i]=max(A[i]-t,0);}if(B[i+1]>0&&A[i]>0){int t=B[i+1];B[i+1]=max(B[i+1]-A[i],0);A[i]=max(A[i]-t,0);}}}cout<<ans<<endl;}return 0;}</span>

C. Given Length and Sum of Digits...

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You have a positive integer m and a non-negative integer s. Your task is to find the smallest and the largest of the numbers that have length m and sum of digits s. The required numbers should be non-negative integers written in the decimal base without leading zeroes.

Input

The single line of the input contains a pair of integers m, s (1 ≤ m ≤ 100, 0 ≤ s ≤ 900) — the length and the sum of the digits of the required numbers.

Output

In the output print the pair of the required non-negative integer numbers — first the minimum possible number, then — the maximum possible number. If no numbers satisfying conditions required exist, print the pair of numbers "-1 -1" (without the quotes).

Sample test(s)

input

2 15

output

69 96

input

3 0

output

-1 -1

贪心加细节

<span style="font-size:18px;">/*************************************************************************    > File Name: cf.cpp    > Author: acvcla    > QQ:     > Mail: [e-mail]     > Created Time: 2014年11月17日 星期一 23时34分13秒 ************************************************************************/#include<iostream>#include<algorithm>#include<cstdio>#include<vector>#include<cstring>#include<map>#include<queue>#include<stack>#include<string>#include<cstdlib>#include<ctime>#include<set>#include<math.h>using namespace std;typedef long long LL;const int maxn = 1e5 + 10;#define rep(i,a,b) for(int i=(a);i<=(b);i++)#define pb push_backint main(){ios_base::sync_with_stdio(false);cin.tie(0);int m,s;char ans1[200],ans2[200];while(cin>>m>>s){bool ok=true;int t1=s/9;for(int i=0;i<150;i++)ans2[i]=ans1[i]='9';if(s==0&&m==1){cout<<"0 0"<<endl;continue;}if(!s&&m>1||s>m*9){cout<<"-1 -1"<<endl;continue;}ans2[m]=ans1[m]=0;int d=s%9;if(d==0){if(t1==m){cout<<ans1<<' '<<ans2<<endl;continue;}ans1[0]='1';ans1[m-t1]='8';for(int i=m-t1-1;i>0;i--)ans1[i]='0';for(int i=t1;i<m;i++)ans2[i]='0';}else{if(t1==m-1){ans1[0]='0'+d;ans2[m-1]='0'+d;}else{ans1[0]='1';ans1[m-t1-1]='0'+d-1;for(int i=m-t1-2;i>0;i--)ans1[i]='0';ans2[t1]='0'+d;for(int i=t1+1;i<m;i++)ans2[i]='0';}}cout<<ans1<<' '<<ans2<<endl;}return 0;}</span>

D. Unbearable Controversy of Being

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Tomash keeps wandering off and getting lost while he is walking along the streets of Berland. It's no surprise! In his home town, for any pair of intersections there is exactly one way to walk from one intersection to the other one. The capital of Berland is very different!

Tomash has noticed that even simple cases of ambiguity confuse him. So, when he sees a group of four distinct intersections a, b, c and d, such that there are two paths from a to c — one through b and the other one through d, he calls the group a "damn rhombus". Note that pairs (a, b), (b, c), (a, d), (d, c) should be directly connected by the roads. Schematically, a damn rhombus is shown on the figure below:

Other roads between any of the intersections don't make the rhombus any more appealing to Tomash, so the four intersections remain a "damn rhombus" for him.

Given that the capital of Berland has n intersections and m roads and all roads are unidirectional and are known in advance, find the number of "damn rhombi" in the city.

When rhombi are compared, the order of intersections b and d doesn't matter.

Input

The first line of the input contains a pair of integers n, m (1 ≤ n ≤ 3000, 0 ≤ m ≤ 30000) — the number of intersections and roads, respectively. Next m lines list the roads, one per line. Each of the roads is given by a pair of integers ai, bi (1 ≤ ai, bi ≤ n;ai ≠ bi) — the number of the intersection it goes out from and the number of the intersection it leads to. Between a pair of intersections there is at most one road in each of the two directions.

It is not guaranteed that you can get from any intersection to any other one.

Output

Print the required number of "damn rhombi".

Sample test(s)

input

5 41 22 31 44 3

output

1

input

4 121 21 31 42 12 32 43 13 23 44 14 24 3

output

12

暴力就好，如果u，v之间长度为2的路径有x条，且x>1,那么显然以u，v为起点和终点的四边形有c(x,2)个

<span style="font-size:18px;">/*************************************************************************    > File Name: cf.cpp    > Author: acvcla    > QQ:    > Mail: [e-mail]     > Created Time: 2014年11月17日 星期一 23时34分13秒 ************************************************************************/#include<iostream>#include<algorithm>#include<cstdio>#include<vector>#include<cstring>#include<map>#include<queue>#include<stack>#include<string>#include<cstdlib>#include<ctime>#include<set>#include<math.h>using namespace std;typedef long long LL;const int maxn = 3e3 + 10;#define rep(i,a,b) for(int i=(a);i<=(b);i++)#define pb push_backint n,m;std::vector<int> G[maxn];int d[maxn][maxn];void dfs(int u,int v,int dist){if(dist>2)return;if(dist==2){++d[u][v];return ;}for(int i=0;i<G[v].size();i++){dfs(u,G[v][i],dist+1);}}int main(){ios_base::sync_with_stdio(false);cin.tie(0);while(cin>>n>>m){for(int i=1;i<=n;i++)G[i].clear();memset(d,0,sizeof d);int u,v;for(int i=1;i<=m;i++){cin>>u>>v;G[u].pb(v);}LL ans=0;for(int i=1;i<=n;i++)dfs(i,i,0);for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(d[i][j]<2||j==i)continue;//cout<<i<<' '<<j<<' '<<d[i][j]<<endl;LL t=d[i][j];ans+=(t*(t-1))/2;}}cout<<ans<<endl;}return 0;}</span>

F. Special Matrices

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

An n × n square matrix is special, if:

• it is binary, that is, each cell contains either a 0, or a 1;
• the number of ones in each row and column equals 2.

You are given n and the first m rows of the matrix. Print the number of special n × n matrices, such that the first m rows coincide with the given ones.

As the required value can be rather large, print the remainder after dividing the value by the given numbermod.

Input

The first line of the input contains three integers n, m, mod (2 ≤ n ≤ 500, 0 ≤ m ≤ n, 2 ≤ mod ≤ 109). Thenm lines follow, each of them contains n characters — the first rows of the required special matrices. Each of these lines contains exactly two characters '1', the rest characters are '0'. Each column of the given m × ntable contains at most two numbers one.

Output

Print the remainder after dividing the required value by number mod.

Sample test(s)

input

3 1 1000011

output

2

input

4 4 1005000110101001011001

output

1

Note

For the first test the required matrices are:

011101110011110101

In the second test the required matrix is already fully given, so the answer is 1.

按行dp，由于每行的和必须为2，所以可以在新建行的时候在列和为1或0的位置上选出两个来添加。直到最终列和全部为2.

具体看代码

<span style="font-size:18px;">/*************************************************************************    > File Name: cf.cpp    > Author: acvcla    > QQ:     > Mail: [e-mail]     > Created Time: 2014年11月17日 星期一 23时34分13秒 ************************************************************************/#include<iostream>#include<algorithm>#include<cstdio>#include<vector>#include<cstring>#include<map>#include<queue>#include<stack>#include<string>#include<cstdlib>#include<ctime>#include<set>#include<math.h>using namespace std;typedef long long LL;const int maxn = 500+10;#define rep(i,a,b) for(int i=(a);i<=(b);i++)#define pb push_backLL d[maxn][maxn],n,m,mod;bool vis[maxn][maxn];int col[maxn];char s[maxn];LL cn2(LL n){return n*(n-1)/2;}LL dfs(LL x,LL y){//当前还有x列为1,y列为0，到达目标状态的方案种数，之所以是每次选2个是因为每行的和都必须为2if(x==0&&y==0)return 1;if(vis[x][y])return d[x][y];d[x][y]=0;vis[x][y]=1;if(x>=2)d[x][y]+=cn2(x)*dfs(x-2,y)%mod;//选出为1的两列让其加上1，此时和为1的为x-2,和为0的为yif(y>=2)d[x][y]+=cn2(y)*dfs(x+2,y-2)%mod;//选出为0的两列让其加上1,此时和为1的为x+2,和为0的为y-2if(x>=1&&y>=1)d[x][y]+=x*y*dfs(x,y-1)%mod;//各选一列加上1，此时和为1的为x,和为0的为y-1return d[x][y]%=mod;}int main(){ios_base::sync_with_stdio(false);cin.tie(0);while(cin>>n>>m>>mod){memset(vis,0,sizeof vis);memset(col,0,sizeof col);rep(i,1,m){cin>>s;for(int j=0;j<n;j++){col[j+1]+=(s[j]=='1');}}LL x=0,y=0;for(int i=1;i<=n;i++){if(col[i]==1)x++;if(col[i]==0)y++;if(col[i]>2){cout<<0<<endl;return 0;}}cout<<dfs(x,y)<<endl;}return 0;}</span>