Eighty seven
Main idea: give n numbers, delete any three numbers and ask whether the remaining 10 numbers add up to 87.
q times, give three numbers at a time, indicating the number deleted.
Because Q is very large, it can be pretreated. 50*50*50
Every time you make a normal backpack, it's n*k*100
bitset optimization considered
//dp[i][j]=dp[i-1][j];
//dp[i][j]=dp[i-1][j-1][k-a[i]]
It can be seen that the current knapsack status can be transmitted by dp[i-1][j] and dp[i-1][j-1](k-a[i]==1), so it can be directly current or on it.
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<cmath>
#include<bitset>
#include<algorithm>
#include<map>
using namespace std;
#define LL long long
#define N 305
#define maxn 200005
#define inf 0x3f3f3f3f
#define sca(x) scanf("%d",&x)
#define fir first
#define sec second
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
int a[55];
int b[55];
bitset<105>dp[2][15];
int mmp[55][55][55];
void init(int x,int y,int z,int n)
{
for(int i=0;i<12;i++)
{
dp[i].reset();
}
dp[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--)
{
//dp[i][j]=dp[i-1][j];
//dp[i][j]=dp[i-1][j-1][k-a[i]]
dp[j] |= (dp[j-1]<<a[i]);
}
}
mmp[x][y][z]=mmp[x][z][y]=(int)(dp[10][87]);
mmp[y][x][z]=mmp[y][z][x]=(int)(dp[10][87]);
mmp[z][y][x]=mmp[z][x][y]=(int)(dp[10][87]);
}
int main()
{
int t;
sca(t);
while(t--)
{
int n;
sca(n);
memset(mmp,0,sizeof(mmp));
for(int i=1;i<=n;i++)sca(a[i]);
int op=0;
for(int i=1;i<=n;i++)
{
for(int j=i;j<=n;j++)
{
for(int k=j;k<=n;k++)
{
init(i,j,k,n);
}
}
}
int q;
sca(q);
while(q--)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
if(mmp[x][y][z])puts("Yes");
else puts("No");
}
}
}