Eighty seven

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 583    Accepted Submission(s): 208

Problem Description
Mr. Fib is a mathematics teacher of a primary school. In the next lesson, he is planning to teach children how to add numbers up. Before the class, he will prepare N cards with numbers. The number on the i-th card is ai. In class, each turn he will remove no more than 3 cards and let students choose any ten cards, the sum of the numbers on which is 87. After each turn the removed cards will be put back to their position. Now, he wants to know if there is at least one solution of each turn. Can you help him?
 

Input
The first line of input contains an integer t (t≤5), the number of test cases. t test cases follow.
For each test case, the first line consists an integer N(N≤50).
The second line contains N non-negative integers a1,a2,...,aN. The i-th number represents the number on the i-th card. The third line consists an integer Q(Q≤100000). Each line of the next Q lines contains three integers i,j,k, representing Mr.Fib will remove the i-th, j-th, and k-th cards in this turn. A question may degenerate while i=j, i=k or j=k.
 

Output
For each turn of each case, output 'Yes' if there exists at least one solution, otherwise output 'No'.
 

Sample Input
1 12 1 2 3 4 5 6 7 8 9 42 21 22 10 1 2 3 3 4 5 2 3 2 10 10 10 10 11 11 10 1 1 1 2 10 1 11 12 1 10 10 11 11 12
 

Sample Output
No No No Yes No Yes No No Yes Yes
 

Source
2016 ACM/ICPC Asia Regional Qingdao Online
 

题意：给你n个数，m次询问，每次询问给你三个数，问在给出的数里面存不存在任意取10个（不包含这三个数）的数的和==87。

做法：利用bitset预处理，bitset是什么玩意？和set一样，都可以看作是集合，只不过里面的值只有0/1，这里面bitset定义为数组，ss[已经用的数字个数][凑出来的和]，那么ss[j]|=ss[j-1]<<num[i];的意思就是在添加了第i个数的时候可以在原有的基础上改变。这样到最后就可以判断出[10][87]是否可达。

#include <iostream>
#include<cstdio>
#include<bitset>
#include<algorithm>
using namespace std;
bitset<90>ss[20];
int n,num[100];
bool dp[55][55][55];
void make(int x,int y,int z)
{
    for(int i=0;i<=11;i++)ss[i].reset();
    ss[0][0]=1;
    for(int i=1;i<=n;i++)
    {
        if(i==x||i==y||i==z||num[i]>87)continue;
        for(int j=10;j>=1;j--)
            ss[j]|=ss[j-1]<<num[i];
    }
    if(ss[10][87]==1)dp[x][y][z]=1;
    else dp[x][y][z]=0;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)scanf("%d",&num[i]);
        for(int i=1;i<=n;i++)
            for(int j=i;j<=n;j++)
                for(int k=j;k<=n;k++)
                    make(i,j,k);
        int m;
        scanf("%d",&m);
        while(m--)
        {
            int xx[3];
            scanf("%d%d%d",&xx[0],&xx[1],&xx[2]);
            sort(xx,xx+3);
            if(dp[xx[0]][xx[1]][xx[2]])puts("Yes");
            else puts("No");
        }
    }
    return 0;
}