Jam‘s balance

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

题意：

Jam有NN个砝码和一个没有游标的天平，现在给他(1 \leq N \leq 20)(1≤N≤20)个砝码，砝码可以放左边，也可以放右边，问可不可以测出所问的重量, 问的个数为(1 \leq M \leq 100)(1≤M≤100)个.

题解：

这道题可以放左边，可以放右边，N=20N=20显然每种状态都枚举是不太现实的，因为每组砝码都可以变成很多种重量，当然也不排除有人乱搞过了这一题，其实这道题是一道贪心的思想，我们看到ww不大，所以可以用0101背包扫一次，当然这还是不够的，这只能放一边，考虑到可以放另一边，就是可以有减的关系，所以反着再背包一遍，注意要判断边界。

#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
const int  N = 5000;
int n,m,a[N+10],dp[N+10],sum;
void DP() {
    memset(dp,0,sizeof(dp));
    dp[0]= 1;
    for(int i = 1; i<= n; i++) {
        for(int k=2;k;k--)
        for(int j = sum*2;j>=a[i];j--) {
            dp[j]|=dp[j-a[i]];
        }
    }
}
int main() {
   int T,x;
   scanf("%d",&T);
   while(T--) {
    scanf("%d",&n);sum=0;
    for(int i = 1;i <= n; i++) scanf("%d", &a[i]),sum+=a[i];
    DP();
    scanf("%d", &m);
    for(int i = 1; i<= m; i++) {
        scanf("%d", &x);
        int g = x+sum&&sum+x>=0&&dp[x+sum];
        if(g)printf("YES\n");
        else printf("NO\n");
    }
   }
    return 0;
}

HDU 5616 Jam's balance 背包DP