NYOJ 325

http://acm.nyist.net/JudgeOnline/problem.php?pid=325

用dp的超时代码

#include<stdio.h>
#include<stdlib.h>
int max(int a,int b)
{
    return a>b?a:b;
}
int dp[201000]={0},a[22]={0};
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
     int sum=0,sum1,r,i,j;
     for(i=0;i<n;i++)
     {
      scanf("%d",&a[i]);
      sum+=a[i];
     }
     sum1=sum/2;
     for(i=0;i<n;i++)
      for(j=sum1;j>=a[i];j--)
      {
        dp[j]=max(dp[j],dp[j-a[i]]+a[i]);
      }
      r=sum-2*dp[sum1];
      printf("%d\n",r);
      
    }
    return 0;
}