HDOJ 5616-Jam's balance【模拟】

Jam's balance

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

Total Submission(s): 188    Accepted Submission(s): 82

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 
    
    
    
    
     
   
   
   
   

 

Source

BestCoder Round #70

 解题思路：

这道题和杭电的1709是一样的，就是 砝码可以放在天枰的左右两端。

#include<stdio.h>  
#include<math.h>  
#include<string.h>
#include<stdlib.h>  
  
int c1[30000],c2[30000];  
int a[30000];  
bool cc[30000];
  
int main()  
{  
    int i,j,n,sum; 
    int nn; 
    scanf("%d",&nn);
    while(nn--)  
    {  
        scanf("%d",&n);
        sum=0;  
        for(i=1;i<=n;i++)  
        {  
            scanf("%d",&a[i]);  
            sum=sum+a[i];  
        }  
        for(i=0;i<=sum;i++)  
        {  
            c1[i]=0;  
            c2[i]=0;  
        }  
        c1[0]=1;  
        for(i=1;i<=n;i++)  
        {  
            for(j=0;j+a[i]<=sum;j++)  
            {  
                if(c1[j]==1)  
                {  
                    c2[j]=1;  
                    c2[j+a[i]]=1;  
                    c2[abs(j-a[i])]=1;  
                }  
            }  
            for(j=0;j<=sum;j++)  
            {  
                c1[j]=c2[j];  
                c2[j]=0;  
            }  
        }  
        int num=0;  
        memset(cc,false,sizeof(cc));
        for(i=1;i<=sum;i++)  
        {  
            if(c1[i]==0)  
            {
                cc[i]=true;
            } 
        }  
        int wnm;
        scanf("%d",&wnm);
        while(wnm--)
        {
            int yyy;
            scanf("%d",&yyy);
            if(yyy>sum)
            {
                printf("NO\n");
                continue;
            }
            if(cc[yyy]==true)
            {
                printf("NO\n");
            }
            else
            {
                printf("YES\n");
            }
        }
    }  
    return 0;  
}