题干：

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 NN cards with numbers. The number on the ii-th card is aiai. In class, each turn he will remove no more than 33 cards and let students choose any ten cards, the sum of the numbers on which is 8787. 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)t (t≤5), the number of test cases. tttest cases follow. 
For each test case, the first line consists an integer N(N≤50)N(N≤50). 
The second line contains NN non-negative integers a1,a2,...,aNa1,a2,...,aN. The ii-th number represents the number on the ii-th card. The third line consists an integer Q(Q≤100000)Q(Q≤100000). Each line of the next QQ lines contains three integers i,j,ki,j,k, representing Mr.Fib will remove the ii-th, jj-th, and kk-th cards in this turn. A question may degenerate while i=ji=j, i=ki=k or j=kj=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

题目大意：

50个数，10W个询问，每次问删掉第i,j,k个数后，是否存在一种选10个数和为87的方案，只需要输出 ’Yes’ 或者 ’No’

解题报告：

   直接可行性背包，然后bitset优化一下就可以了。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#include<bitset>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
bitset<111> bs[111];
int a[MAX],n;
bool dp[111][111][111];
bool ok(int x,int y,int z) {for(int i = 1; i<=n; i++) bs[i].reset();bs[0][0]=1;for(int i = 1; i<=n; i++) {if(i == x || i == y || i == z) continue;for(int j = 10; j>=1; j--) {bs[j] |= (bs[j-1] << a[i]);}}return bs[10][87];
}
int main()
{int t;cin>>t;while(t--) {scanf("%d",&n);memset(dp,0,sizeof dp);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++) {dp[i][j][k]=0;if(ok(i,j,k)) dp[i][j][k] = 1;}}}int q,x[4];scanf("%d",&q);while(q--) {for(int i = 1; i<=3; i++) scanf("%d",x+i);sort(x+1,x+4);if(dp[x[1]][x[2]][x[3]] == 1) puts("Yes");else puts("No");}}return 0 ;
}