HDU 5616 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

121 43245

 

Sample Output

NOYESYES
Hint
For the Case 1:Put the 4 weight aloneFor the Case 2:Put the 4 weight and 1 weight on both side

题意：有一个没有游标的天平，和n个秤砣，m个询问， 每次一个k，
问可否秤出k这个重量。 秤砣可以放两边。

dp[n][sum]存储可达与不可达....
dp[i][j] 表示放前i个秤砣能否称出j重量 
因为相减可能为负 所以把全部值加上sum 避免出现负数
因此初始为 dp[0][sum] = 1 

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <queue>#include <vector>#include <cmath>#include <stack>#include <string>#include <map>#include <set>#define pi acos(-1)#define LL long long#define ULL unsigned long long#define inf 0x3f3f3f3f#define INF 1e18#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1using namespace std;typedef pair<int, int> P;const int maxn = 1e5 + 5;const int mod = 1e8;int dp[25][maxn];int a[25];int main(void){//freopen("C:\\Users\\wave\\Desktop\\NULL.exe\\NULL\\in.txt","r", stdin);    int T, i, j, n, m, sum, t;    cin >> T;    while (T--)    {        sum = 0;        cin >> n;        for (i = 1; i <= n; i++){            cin >> a[i];            sum += a[i];        }        memset(dp, 0, sizeof(dp));        dp[0][sum] = 1;        for (i = 1; i <= n; i++){            for (j = 1; j <= 2*sum; j++){                if (dp[i-1][j]){                    dp[i][j] = 1;                    dp[i][j - a[i]] = 1;                    dp[i][j + a[i]] = 1;                }            }        }        cin >> m;        while (m--){            cin >> t;            if (t > sum) puts("NO");            else if (dp[n][t + sum])                puts("YES");            else puts("NO");        }    }return 0;}