Eighty seven

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)

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 Ncards 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

bitset把87*n优化成10*n；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
const int N=1e5+10,M=1e6+10,inf=1e9+10,mod=1e9+7;
const ll INF=1e18+10;
bitset<90>dp[11];
int ans[60][60][60];
int a[100],n,m;
int q[5];
int check(int x,int y,int z)
{
    for(int i=0;i<=n;i++)dp[i].reset();
    dp[0][0]=1;
    for(int i=1;i<=n;i++)
    {
        if(i!=x&&i!=y&&i!=z)
            for(int t=10;t>=1;t--)
                dp[t]|=dp[t-1]<<a[i];
    }
    if(dp[10][87]==1)
        return 1;
        return 0;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(ans,0,sizeof(ans));
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
            for(int j=i;j<=n;j++)
                for(int k=j;k<=n;k++)
                if(check(i,j,k))ans[i][j][k]=1;
        scanf("%d",&m);
        while(m--)
        {
            for(int i=0;i<3;i++)
                scanf("%d",&q[i]);
            sort(q,q+3);
            if(ans[q[0]][q[1]][q[2]])
                printf("Yes\n");
            else
                printf("No\n");
        }
    }
    return 0;
}