Codeforces Round #510 (Div. 2) A&B By cellur925

第一次CF祭==

由于太菜了只做了前两题==

因为在第一题上耗费时间太多了，我还是太菜了==。

A. Benches

time limit per test

1 second

memory limit per test

256 megabytes

There are nn benches in the Berland Central park. It is known that aiai people are currently sitting on the ii-th bench. Another mm people are coming to the park and each of them is going to have a seat on some bench out of nn available.

Let kk be the maximum number of people sitting on one bench after additional mm people came to the park. Calculate the minimum possible kk and the maximum possible kk.

Nobody leaves the taken seat during the whole process.

Input

The first line contains a single integer nn (1≤n≤100)(1≤n≤100) — the number of benches in the park.

The second line contains a single integer mm (1≤m≤10000)(1≤m≤10000) — the number of people additionally coming to the park.

Each of the next nn lines contains a single integer aiai (1≤ai≤100)(1≤ai≤100) — the initial number of people on the ii-th bench.

Output

Print the minimum possible kk and the maximum possible kk, where kk is the maximum number of people sitting on one bench after additional mm people came to the park.

一句话题意：对一个确定的数列增加元素，求增加元素后序列最大值k的最大值和最小值

观察样例发现，k的最大值一定是把所有元素加到数列中最大值的结果。

那么最小值呢？几乎40分钟都在思考这个，从二分想到平均数云云，都不对，后来突然想到，我们从宏观来审视这个问题，因为增加的元素是全部都要进数列的，所以我们求出最后数列的全部元素总和，除以数列项数即可。（看起来很小学奥数的算法结果想了很久。。。）

于是愉快的交，几分钟后就被hack了，非常不爽...原因写在了注释里。还是考虑情况不全面。

 #include<cstdio>
 #include<algorithm>

 using namespace std;

 int n,m,maxans,sum,minans;
 int a[];

 int main()
 {
     scanf("%d%d",&n,&m);
     for(int i=;i<=n;i++)
     {
         scanf("%d",&a[i]);
         maxans=max(maxans,a[i]);
         sum+=a[i];
     }
     sum+=m;
     minans=max(maxans,(sum+n-)/n);
     //结果肯定不能比maxans小
     // 面对不能整除的情况我们+1，然后为了让结果具有普适性，所以+n/n，但是又怕在整除的时候加多，所以-1
     printf("%d %d",minans,maxans+m);
     return ;
 }

B. Vitamins

time limit per test

2 seconds

memory limit per test

256 megabytes

Berland shop sells nn kinds of juices. Each juice has its price cici. Each juice includes some set of vitamins in it. There are three types of vitamins: vitamin "A", vitamin "B" and vitamin "C". Each juice can contain one, two or all three types of vitamins in it.

Petya knows that he needs all three types of vitamins to stay healthy. What is the minimum total price of juices that Petya has to buy to obtain all three vitamins? Petya obtains some vitamin if he buys at least one juice containing it and drinks it.

Input

The first line contains a single integer nn (1≤n≤1000)(1≤n≤1000) — the number of juices.

Each of the next nn lines contains an integer cici (1≤ci≤100000)(1≤ci≤100000) and a string sisi — the price of the ii-th juice and the vitamins it contains. String sisi contains from 11 to 33 characters, and the only possible characters are "A", "B" and "C". It is guaranteed that each letter appears no more than once in each string sisi. The order of letters in strings sisi is arbitrary.

Output

Print -1 if there is no way to obtain all three vitamins. Otherwise print the minimum total price of juices that Petya has to buy to obtain all three vitamins.

本题显然是背包类的dp了，但是我开始竟然写了模拟（？？？），后来发现7种情况可能不全有。。我太菜了。

看到两种好的做法：Chemist的，码量较大，但是很容易理解。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>

 using namespace std;

 int n;
 int price[],val[];
 char Chemist[];
 int dp[][][][];

 int main()
 {
     scanf("%d",&n);
     for(int i=;i<=n;i++)
     {
         scanf("%d",&price[i]);
         scanf("%s",Chemist+);
         for(int j=;j<=strlen(Chemist+);j++)
         {
             if(Chemist[j]=='A') val[i]+=;
             if(Chemist[j]=='B') val[i]+=;
             if(Chemist[j]=='C') val[i]+=;
         }
     }
     memset(dp,,sizeof(dp));
     for(int i=;i<=n;i++)
         dp[i][][][]=;//两个初值操作，必须有。
     for(int i=;i<=n;i++)
     {
         for(int j=;j<=;j++)
             for(int l=;l<=;l++)//继承之上，必须有。
                 for(int r=;r<=;r++)
                     dp[i][j][l][r]=dp[i-][j][l][r];
         if(val[i]==)
         {
             for(int j=;j<=;j++)
                 for(int l=;l<=;l++)
                     for(int r=;r<=;r++)
                         dp[i][][][]=min(dp[i-][j][l][r]+price[i],dp[i][][][]);
         }
         else if(val[i]==)
         {//VITAMIN A可能有或没有 分开考虑
             for(int j=;j<=;j++)
                 for(int l=;l<=;l++)//从上一个转移来，就是带上当前的了，也就是决策
                     dp[i][][][]=min(dp[i-][][j][l]+price[i],dp[i][][][]);
             for(int j=;j<=;j++)
                 for(int l=;l<=;l++)
                     dp[i][][][]=min(dp[i-][][j][l]+price[i],dp[i][][][]);
         }
         else if(val[i]==)
         {
             for(int j=;j<=;j++)
                 for(int l=;l<=;l++)
                     dp[i][][][]=min(dp[i-][j][][l]+price[i],dp[i][][][]);
             for(int j=;j<=;j++)
                 for(int l=;l<=;l++)
                     dp[i][][][]=min(dp[i-][j][][l]+price[i],dp[i][][][]);
         }
         else if(val[i]==)
         {
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
         }
         else if(val[i]==)
         {
             for(int j=;j<=;j++)
                 for(int l=;l<=;l++)
                     dp[i][][][]=min(dp[i-][j][l][]+price[i],dp[i][][][]);
             for(int j=;j<=;j++)
                 for(int l=;l<=;l++)
                     dp[i][][][]=min(dp[i-][j][l][]+price[i],dp[i][][][]);
         }
         else if(val[i]==)
         {
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
         }
         else if(val[i]==)
         {
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
             dp[i][][][]=min(dp[i][][][],dp[i-][][][]+price[i]);
         }
     }
     int ans=1e9;
     for(int i=;i<=n;i++)
     //从每一个地方寻找！！ 因为决策可能不在最后
         ans=min(ans,dp[i][][][]);
     if(ans==1e9) printf("-1");
     else printf("%d",ans);
     return ;
 }

L_A的，精简代码，巧妙运用位运算，神仙做法。

 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 using namespace std;

 int i,j,n,f[],c,t,len;
 char vt[];

 int main()
 {
     scanf("%d",&n);
     memset(f,0x3f,sizeof(f));
     f[]=;
     for(i=;i<=n;++i)
     {
         scanf("%d%s",&c,vt+);
         len=strlen(vt+);
         t=;
         for(j=;j<=len;++j)
             switch(vt[j])
             {
                 case 'A':
                     t|=;
                     break;
                 case 'B':
                     t|=;
                     break;
                 case 'C':
                     t|=;
                     break;
             }
         for(j=;j<=;++j)
             f[t|j]=min(f[t|j],f[j]+c);
     }
     if(f[]==0x3f3f3f3f) printf("-1");
     else printf("%d",f[]);
     return ;
 }

二位神仙都用了1,2,4这类二进制表示的方法，我开始也想用的，只不过用了1,2,3就会gg了，因为相加后可能会有重复，用2^n的数字表示，肥肠好。