HDOJ 5616 Jam's balance(母函数)

Jam's balance

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 109    Accepted Submission(s): 47

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：有n个砝码和一个没有游标的天平，砝码能放左边和右边。给出m个重量ki，问这些ki哪些是能测出来的，哪些是不能测出来的？

题解：妹妹的！！！无语啊，k没给范围，完了特判k>sum直接输出NO。内hack了。  母函数啦，前天才做的题。

链接： HDOJ 1709的题解  一模一样的题啦。

代码如下：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
int c1[2010],c2[2010];
int a[22];
int main()
{
	int t,n,m,i,j,k,sum,num;
	scanf("%d",&t);
	while(t--)
	{
		sum=0;
		scanf("%d",&n);
		for(i=1;i<=n;++i)
		{
			scanf("%d",&a[i]);
			sum+=a[i];
		}
		memset(c1,0,sizeof(c1));
		memset(c2,0,sizeof(c2));
		c1[0]=c1[a[1]]=1;
		num=a[1];	
		for(i=2;i<=n;++i)
		{
			for(j=0;j<=num;++j)
			{
				for(k=0;k<=a[i]&&k+j<=sum;k+=a[i])
				{
					c2[k+j]+=c1[j];
					c2[abs(j-k)]+=c1[j];
				}
			}
			num+=a[i];
			for(j=0;j<=num;++j)
			{
				c1[j]=c2[j];
				c2[j]=0;
			}
		}
		scanf("%d",&m);
		while(m--)
		{
			scanf("%d",&num);
			if(num>sum)
			{
				printf("NO\n");
				continue;
			} 
			if(c1[num])
				printf("YES\n");
			else
				printf("NO\n");
		}
	}
	return 0;
}