杭电 hdu 5616 Jam's balance【思维】【暴力枚举】

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

因为N不大，并且wi也不大，我们这里可以应用数组标记的方法，暴力枚举出每一种可能。

首先我们需要两个vis数组 。我们这里规定1代表有这个可能，0代表没有这个可能、这里对应核心代码详解：
并且这里要注意，题干中样例如果加一个3的话，也是应该输出YES的~

        memset(vis1,0,sizeof(vis1));
        memset(vis2,0,sizeof(vis2));
        vis1[0]=1;//0是一定可以称出来的
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<=2000;j++)//20*100是最大的边界
            {
                if(vis1[j]!=0)//如果vis1数组上标记过这个数。
                {
                    vis2[j]=1;
                    vis2[j+a[i]]=1;//处理+
                    vis2[abs(j-a[i])]=1;//处理-
                }
            }
            for(int k=0;k<=2000;k++)//数组赋值过去
            {
                vis1[k]=vis2[k];
                vis2[k]=0;
            }
        }

蓝后是完整的AC代码：

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

using namespace std;
int vis1[2005];
int vis2[2005];
int a[2005];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        memset(vis1,0,sizeof(vis1));
        memset(vis2,0,sizeof(vis2));
        vis1[0]=1;
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<=2000;j++)
            {
                if(vis1[j]!=0)
                {
                    vis2[j]=1;
                    vis2[j+a[i]]=1;
                    vis2[abs(j-a[i])]=1;
                }
            }
            for(int k=0;k<=2000;k++)
            {
                vis1[k]=vis2[k];
                vis2[k]=0;
            }
        }
        int m;
        scanf("%d",&m);
        while(m--)
        {
            long long int q;
            scanf("%I64d",&q);
            if(vis1[q]==0||q<0||q>2000)
            {
                printf("NO\n");
            }
            else
            {
                printf("YES\n");
            }
        }
    }
}