Jam's balance

Jam's balance

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2915    Accepted Submission(s): 1121

Problem Description

Jim has a balance and N weights. (1≤N≤20)
The balance can only tell whether things on different side are the same weight.
Weights can be put on left side or right side arbitrarily.
Please tell whether the balance can measure an object of weight M.

 

Input

The first line is a integer T(1≤T≤5), means T test cases.
For each test case :
The first line is N, means the number of weights.
The second line are N number, i'th number wi(1≤wi≤100) means the i'th weight's weight is wi.
The third line is a number M. M is the weight of the object being measured.

 

Output

You should output the "YES"or"NO".

 

Sample Input

1 2 1 4 3 2 4 5

 

Sample Output

NO YES YES

Hint

For the Case 1:Put the 4 weight alone For the Case 2:Put the 4 weight and 1 weight on both side

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
const int bit = (1 << 10);
const int oo = 6e3 + 10;

int T;
int n, Query, c[maxn], dp[maxn];

int main() {
#ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
#endif
    scanf("%d", &T);
    while (T--) {
        scanf("%d", &n);
        for (register int i = 1; i <= n; ++i) {
            scanf("%d", &c[i]);
        }
        sort(c + 1, c + 1 + n);
        memset(dp, 0, sizeof(dp));
        dp[0] = 1;
        for (register int i = 1; i <= n; ++i) {
            for (register int j = oo; j >= c[i]; --j) {
                dp[j] = dp[j - c[i]];
            }
        }
        for (register int i = 1; i <= n; ++i) {
            for (register int j = c[i]; j <= oo; ++j) {
                dp[j - c[i]] |= dp[j];
            }
        }
        scanf("%d", &Query);
        int cur;
        while (Query--) {
            scanf("%d", &cur);
            if (dp[cur]) {
                puts("YES");
            } else {
                puts("NO");
            }
        }
    }
    return 0;
}