Codeforces Round #436 (Div. 2)

A , B , C , D , E

F 有时间更新额 （感觉自己懒死了 ， 逃~~）

算了 把E题先放到前面来吧
忘记看大神的写法了。。。。有时间更新
Talk is cheap , show me the code (初中英语水平~~逃)

E. Fire

Polycarp is in really serious trouble — his house is on fire! It’s time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di — the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti ≥ di, then i-th item cannot be saved.

Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b — in ta + tb seconds after fire started.
Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of items in Polycarp’s house.

Each of the following n lines contains three integers ti, di, pi (1 ≤ ti ≤ 20, 1 ≤ di ≤ 2 000, 1 ≤ pi ≤ 20) — the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output

In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m — the number of items in the desired set. In the third line print m distinct integers — numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.

题意：救火，n件物品（t,d,p）救它要t，在d之后就不用救了（已经烧了），价值P，求拯救的最大值和拯救的次序
大致一看，如果不考虑d的因素，就是一道背包DP 。
考虑已知答案，求次序：是贪心的做法，按照d的大小排序，贪心先救最先损坏的（这也是人之常情）
所以按照d排序后，用背包DP的算法，之后根据最大值，DFS（伪DFS，~~~）

/*******************************************************************  author : ********* problem : #    time : 2017-09-26 21:11:26 *******************************************************************/#include <iostream>#include <fstream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <time.h>#include <limits.h>#include <assert.h>#include <set>#include <map>#include <stack>#include <queue>#include <list>#include <bitset>#include <vector>using namespace std;#define forn(i, n) for (int i = 0; i < (int)(n); ++i)#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int>#define pi pair<int,int>#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<" "; cout<<a[n]<<endl;#define lowbit(x) ((x)&-(x))#define SI(a) ((a).size())#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define MAX 35#define INF (0x3f3f3f3f)#define F (1000000007)#define LL long longconst int N = 2005;int f[105][N];struct WHATA {int t, d, p, n;} WTF[105];bool cmp(WHATA a, WHATA b) {return a.d < b.d;}vector<int>v;void DFS(int s, int d, int p) {    // printf("s = %d d = %d p = %d \n",s,d,p);    if (s == 0 || p == 0) return;    if(f[s-1][d] == p) DFS(s-1,d,p);    else {        RepD(i,d-WTF[s].t) {            // printf("f[%d][%d] = %d wtf[%d].p = %d p = %d\n"                // ,s-1,i,f[s-1][i],s,WTF[s].p,p);            if(f[s-1][i] + WTF[s].p == p) {                DFS(s-1,i,p-WTF[s].p);                break;            }        }        v.pb(s);    }}int main(int argc, char const *argv[]){    // freopen("data.in", "r", stdin);    int n; scanf("%d", &n);    For(i, n) scanf("%d%d%d", &WTF[i].t, &WTF[i].d, &WTF[i].p), WTF[i].n = i;    sort(WTF + 1, WTF + n + 1, cmp);    // Rep(i,n) printf("%d\n",WTF[i].d);    For(i, n) {        Rep(j, N) {            if (j - WTF[i].t >= 0 && j < WTF[i].d)                f[i][j] = max(f[i - 1][j], f[i - 1][j - WTF[i].t] + WTF[i].p);            else f[i][j] = f[i - 1][j];        }    }    int pos , maxx = 0;    Rep(i, N) if (maxx < f[n][i]) maxx = f[n][i] , pos = i;    DFS(n, pos, maxx);    printf("%d\n%d\n",maxx,v.size());    Rep(i,v.size()) printf(i==v.size()-1?"%d\n":"%d ",WTF[v[i]].n);    return 0;}

A. Fair Game

Petya and Vasya decided to play a game. They have n cards (n is an even number). A single integer is written on each card.

Before the game Petya will choose an integer and after that Vasya will choose another integer (different from the number that Petya chose). During the game each player takes all the cards with number he chose. For example, if Petya chose number 5 before the game he will take all cards on which 5 is written and if Vasya chose number 10 before the game he will take all cards on which 10 is written.

The game is considered fair if Petya and Vasya can take all n cards, and the number of cards each player gets is the same.

Determine whether Petya and Vasya can choose integer numbers before the game so that the game is fair.
Input

The first line contains a single integer n (2≤n≤100) — number of cards. It is guaranteed that n is an even number.

The following n lines contain a sequence of integers a1,a2,…,an (one integer per line, 1≤ai≤100) — numbers written on the n cards.
Output

If it is impossible for Petya and Vasya to choose numbers in such a way that the game will be fair, print “NO” (without quotes) in the first line. In this case you should not print anything more.

In the other case print “YES” (without quotes) in the first line. In the second line print two distinct integers — number that Petya should choose and the number that Vasya should choose to make the game fair. If there are several solutions, print any of them.

A 题照例水题 ， 直接统计就好了
题意：给你一个数组，问能否分成两个长度相等，里面元素都一样的子集
强行使用STL

/*******************************************************************  author : ******* problem : A    time : 2017-09-26 18:52:08 *******************************************************************/#include <iostream>#include <fstream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <time.h>#include <limits.h>#include <assert.h>#include <set>#include <map>#include <stack>#include <queue>#include <list>#include <bitset>#include <vector>using namespace std;#define forn(i, n) for (int i = 0; i < (int)(n); ++i)#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int>#define pi pair<int,int>#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<" "; cout<<a[n]<<endl;#define lowbit(x) ((x)&-(x))#define SI(a) ((a).size())#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define MAX 35#define INF (0x3f3f3f3f)#define F (1000000007)#define LL long longint main(int argc, char const *argv[]){    ios::sync_with_stdio(false);// you may need this    // fstream cin("data.in");    // freopen("data.in", "r", stdin);    int n;cin>>n;    std::vector<int> arr(105,0);    std::vector<int> brr;    Rep(i,n) {        int x ; cin>>x;        if(arr[x]==0) brr.pb(x);        arr[x]++;    }    if(brr.size()!=2 || arr[brr[0]] != arr[brr[1]]) return 0*printf("NO\n");    sort(brr.begin(), brr.end());    cout<<"YES"<<endl;    cout<<brr[0]<<" "<<brr[1]<<endl;    return 0;}

B. Polycarp and Letters

Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string s consisting only of lowercase and uppercase Latin letters.

Let A be a set of positions in the string. Let’s call it pretty if following conditions are met:

letters on positions from A in the string are all distinct and lowercase;there are no uppercase letters in the string which are situated between positions from A (i.e. there is no such j that s[j] is an uppercase letter, and a1 < j < a2 for some a1 and a2 from A). 

Write a program that will determine the maximum number of elements in a pretty set of positions.
Input

The first line contains a single integer n (1 ≤ n ≤ 200) — length of string s.

The second line contains a string s consisting of lowercase and uppercase Latin letters.
Output

Print maximum number of elements in pretty set of positions for string s.

B 题 感觉也好水 的说
题意：给你一个字符串，对其中每个连续小写字母组成的串，问其中串中不同字母数的最大值是多少？
直接模拟就好了

/*******************************************************************  author : ******** problem : #    time : 2017-09-26 19:06:25 *******************************************************************/#include <iostream>#include <fstream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <time.h>#include <limits.h>#include <assert.h>#include <set>#include <map>#include <stack>#include <queue>#include <list>#include <bitset>#include <vector>using namespace std;#define forn(i, n) for (int i = 0; i < (int)(n); ++i)#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int>#define pi pair<int,int>#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<" "; cout<<a[n]<<endl;#define lowbit(x) ((x)&-(x))#define SI(a) ((a).size())#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define MAX 35#define INF (0x3f3f3f3f)#define F (1000000007)#define LL long longint main(int argc, char const *argv[]){    ios::sync_with_stdio;    // fstream cin("data.in");    // freopen("data.in", "r", stdin);    int len;cin>>len;    string s;cin>>s;        int now = 0 , maxx = 0;    bool lowletter[26] = {false};    Rep(i,len) {;        if(s[i]<='z' && s[i]>='a') {            if(lowletter[s[i] - 'a'] == false) now++ , lowletter[s[i]-'a'] = true;        }        else  {            maxx = max(maxx,now);            now = 0;            MEM(lowletter);        }    }    maxx = max(maxx,now);    cout<<maxx<<endl;    return 0;}

C. Bus

A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.

The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.

There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.

What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input

The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output

Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.

题意：公交车从0到a 在f加油，油箱中有b ，往返算两次，问k次最少加多次油

感觉弄了个傻傻的写法，直接模拟，每次到加油站计算剩下的油是否够到下次加油站。
如果够，就不加；否则 就加；

/*******************************************************************  author : ******** problem : #    time : 2017-09-26 19:23:18 *******************************************************************/#include <iostream>#include <fstream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <time.h>#include <limits.h>#include <assert.h>#include <set>#include <map>#include <stack>#include <queue>#include <list>#include <bitset>#include <vector>using namespace std;#define forn(i, n) for (int i = 0; i < (int)(n); ++i)#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int>#define pi pair<int,int>#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<" "; cout<<a[n]<<endl;#define lowbit(x) ((x)&-(x))#define SI(a) ((a).size())#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define MAX 35#define INF (0x3f3f3f3f)#define F (1000000007)#define LL long longstruct Car{    int soil;}car;int main(int argc, char const *argv[]){    ios::sync_with_stdio;    // fstream cin("data.in");    int a,b,f,k;cin>>a>>b>>f>>k;    car.soil = b ;    LL ans = 0 ;    For(i,k) {        if(i==1) car.soil -= f;        if(car.soil<0) {            ans = -1;break;        }        int ll = i&1 ? 2*(a-f) : 2*f;        if(i==k) ll = i&1? a-f : f;        // printf("i = %d ll = %d left = %d \n",i,ll,car.soil);        if(ll>car.soil) {            car.soil = b ;            ans++;        }        if(car.soil<ll) {            ans = -1;            break;        }        car.soil-=ll;    }    printf("%lld\n",ans );    return 0;}

D. Make a Permutation!

Ivan has an array consisting of n elements. Each of the elements is an integer from 1 to n.

Recently Ivan learned about permutations and their lexicographical order. Now he wants to change (replace) minimum number of elements in his array in such a way that his array becomes a permutation (i.e. each of the integers from 1 to n was encountered in his array exactly once). If there are multiple ways to do it he wants to find the lexicographically minimal permutation among them.

Thus minimizing the number of changes has the first priority, lexicographical minimizing has the second priority.

In order to determine which of the two permutations is lexicographically smaller, we compare their first elements. If they are equal — compare the second, and so on. If we have two permutations x and y, then x is lexicographically smaller if xi < yi, where i is the first index in which the permutations x and y differ.

Determine the array Ivan will obtain after performing all the changes.
Input

The first line contains an single integer n (2≤n≤200000) — the number of elements in Ivan’s array.

The second line contains a sequence of integers a1,a2,…,an (1≤ai≤n) — the description of Ivan’s array.
Output

In the first line print q — the minimum number of elements that need to be changed in Ivan’s array in order to make his array a permutation. In the second line, print the lexicographically minimal permutation which can be obtained from array with q changes.

D 题 本质上是不难的 ， 主要是怎么写的简单~
题意：对给定数组，最少修改多少个数字，使数组变成1~n的全排列，输出最少改变下字典序最小的答案
对可以修改的位置判断是否修改，（感觉说的好简单）

/*******************************************************************  author : ******** problem : #    time : 2017-09-26 19:50:02 *******************************************************************/#include <iostream>#include <fstream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <time.h>#include <limits.h>#include <assert.h>#include <set>#include <map>#include <stack>#include <queue>#include <list>#include <bitset>#include <vector>using namespace std;#define forn(i, n) for (int i = 0; i < (int)(n); ++i)#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int>#define pi pair<int,int>#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<" "; cout<<a[n]<<endl;#define lowbit(x) ((x)&-(x))#define SI(a) ((a).size())#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define MAX 35#define INF (0x3f3f3f3f)#define F (1000000007)#define LL long longint main(int argc, char const *argv[]){    // ios::sync_with_stdio;    fstream cin("data.in");    int n;cin>>n;    std::vector<int> vis(n+1,0);    std::vector<int> ans;    ans.pb(0);    std::vector<int> arr;    std::vector<int> mark(n+1,0);    For(i,n) {        int x;cin>>x;        ans.pb(x);        vis[x]++;    }    For(i,n) {        if(vis[i]==0) arr.pb(i);    }    sort(arr.begin(), arr.end());    int tot = 0;    printf("%d\n",arr.size());    // Rep(i,arr.size()) printf(i==arr.size()-1?"%d\n":"%d ",arr[i]);    // For(i,n) printf(i==n?"%d\n":"%d ",ans[i]);    For(i,n) {        int tmp;        if(vis[ans[i]]==1 && mark[ans[i]]==0) tmp = ans[i];        else if(mark[ans[i]]==0 && ans[i]<arr[tot]) {            tmp = ans[i];            mark[ans[i]] = 1;            vis[ans[i]]--;                  }        else {            tmp = arr[tot++];            vis[ans[i]]--;        }        printf(i==n ?"%d\n":"%d " ,tmp);    }    return 0;}