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

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
 

 

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;  }