母函数解决 Jam's balance hdu5616

Jam's balance

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1572    Accepted Submission(s): 669

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

 

题母的 意思是  给 几个砝码    砝码 可以放在天平的 两边 或者一边 ，  问你 给你几个 测试数据 砝码 能否 恰好 称出 这个数据 

这个 题 解法有多种  ， 我用的母函数 ： 

模板一：

//1.0：母函数 解决 模板一 

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

#define abs(x) ((x)<0? -(x):(x))// - 一定要在（）外 
int a[2015],b[2015];
int v[2015];
int main()
{
	int T,n,m;
	int i,j,k,kk;
	while(~scanf("%d",&T))
	{
		while(T--)
		{
			int sum=0;
			memset(v,0,sizeof(v));//清零操作
			memset(a,0,sizeof(a));
			memset(b,0,sizeof(b));
			scanf("%d",&n);
			for(i=1;i<=n;i++)
			{
				scanf("%d",&v[i]);
				sum+=v[i];
			}
			a[v[1]]=a[0]=1;//初始 化  
			for(i=2;i<=n;i++)//次数 
			{
				memset(b,0,sizeof(b)); //╮(╯_╰)╭进行一次 就  初始为0 
				for(j=0;j<=sum;j++)// 每个 v[]  
				{
					for(k=0;k+j<=sum&&k<=v[i];k+=v[i])//每个a 
					{
						b[k+j]+=a[j];//天平一边 
						b[abs(k-j)]+=a[j];//天平两边 
					}
				}
				memcpy(a,b,sizeof(b));	
			}	
			scanf("%d",&m);
//			printf("abs=%d\n",abs(5-6));
//			for(i=0;i<sum;i++)
//				{
//					printf("%d \n",a[i]);
//				
//				}
			while(m--)
			{
				scanf("%d",&kk);
				if(a[kk])
				{
					printf("YES\n");
				}
				else
				{
					printf("NO\n");
				}
			}
		
		}
	}
	return 0;
}