hdu 5890 Eighty seven

Eighty seven

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 243    Accepted Submission(s): 74

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

1121 2 3 4 5 6 7 8 9 42 21 22101 2 33 4 52 3 210 10 1010 11 1110 1 11 2 101 11 121 10 1011 11 12

 

Sample Output

NoNoNoYesNoYesNoNoYesYes

 

Source

2016 ACM/ICPC Asia Regional Qingdao Online

 

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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。

思路：用bitset暴力dp，时间卡着过，暴力打表用dp[i][j][k]表示存不存在，然后就是用bitset优化0 1背包。剩下的看代码：

#include<set>#include<map>#include<ctime>#include<cmath>#include<stack>#include<queue>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<bitset>#include<functional>typedef long long LL;using namespace std;#define inf 0x3f3f3f3f#define maxn 55typedef long long LL;int a[maxn], dp[maxn][maxn][maxn];bitset<90>s[11];int n;bool solve(int x, int y, int z){for (int i = 0; i<11; i++)s[i].reset();s[0][0] = 1;for (int i = 1; i <= n; i++){if (i == x || i == y || i == z || a[i]>87)continue;for (int j = 10; j>0; j--){s[j] |= s[j - 1] << a[i];}}if (s[10][87])return 1;elsereturn 0;}int main(){int t;scanf("%d", &t);while (t--){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++){dp[i][j][k] = solve(i, j, k);}}}int q;scanf("%d", &q);while (q--){int b[3];scanf("%d%d%d", &b[0], &b[1], &b[2]);sort(b, b + 3);if (dp[b[0]][b[1]][b[2]])printf("Yes\n");elseprintf("No\n");}}}