链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=456
价值与重量相等的01背包..背包的容量为邮票总和的一半
#include <iostream> #include<cmath> #include<cstdio> #include<cstring> using namespace std; int dp[500005]; int v[1005]; int main() //价值与总量相等的01背包 { //freopen("1.txt","r",stdin); int m,n; int i,j; int sum; int tv; cin>>m; while(m--) { cin>>n; sum=0; for(i=0;i<n;i++) { cin>>v[i]; sum+=v[i]; } memset(dp,0,sizeof(dp)); tv=sum/2; for(i=0;i<n;i++) for(j=tv;j>=v[i];j--) dp[j]= max( dp[j], dp[j-v[i]]+v[i]); cout<<int(sum-dp[tv]*2)<<endl; } return 0; }