UVA - 562 Dividing coins （01背包）

题目大意：给一串数字作为硬币的币值，将这些硬币分给两个人A和B，要求越平均越好。
解析：01背包问题，先计算出总的钱数sum，让后在不超过sum/2的条件下把钱分给A，那么A和B的相差则为 sum - 2*A
因为硬币的价格即为它的体积，
则状态转移方程为：dp[i][j] = max(dp[i][j],dp[i-1][j - coin[i]] + coin[i]);

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 105;
const int MAX = 50010;
int coin[N] ,dp[N][MAX];
int main() {
	int t, n;
	scanf("%d",&t);
	while(t--) {
		scanf("%d",&n);
		int sum = 0;
		for(int i = 1; i <= n; i++) {
			scanf("%d",&coin[i]);
			sum += coin[i];
		}
		memset(dp,0,sizeof(dp));
		for(int i = 1; i <= n; i++) {
			for(int j = 0; j <= sum/2; j++) {
				dp[i][j] = dp[i-1][j];
				if(j >= coin[i]) {
					dp[i][j] = max(dp[i][j],dp[i-1][j - coin[i]] + coin[i]);
				}
			}
		}
		printf("%d\n",sum - 2*dp[n][sum/2]);
	}
	return 0;
}