HDOJ 5616-Jam's balance【模拟】

水深无声 2022-08-20 04:24 17阅读 0赞

# 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)![Image 1][]  
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)![Image 1][], means T test cases.  
For each test case :  
The first line is  N![Image 1][], means the number of weights.  
The second line are  N![Image 1][] number, i'th number  w![Image 1][]i![Image 1][](1≤w![Image 1][]i![Image 1][]≤100)![Image 1][] means the i'th weight's weight is  w![Image 1][]i![Image 1][]![Image 1][].  
The third line is a number  M![Image 1][].  M![Image 1][] 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][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;  
    }

[Image 1]: 
[BestCoder Round _70]: http://acm.hdu.edu.cn/search.php?field=problem&key=BestCoder+Round+%2370&source=1&searchmode=source