HUD-1171 Big Event in HDU（多重背包）

Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don’t know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0<N<1000) kinds of facilities (different value, different kinds).
中文总结：分割，使的两部分的价值平均。若不能平均，要求差最小且大数在前
输入
Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 – the total number of different facilities). The next N lines contain an integer V (0<V<=50 --value of facility) and an integer M (0<M<=100 --corresponding number of the facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.
输出
For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.
样例输入
2
10 1
20 1
3
10 1
20 2
30 1
-1
样例输出
20 10
40 40
注意
一个无聊的坑
本题当n为负值时结束，不是-1或0.

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#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<string>
#define ll long long
#define dd double
using namespace std;

struct bag {
	ll x, y, z;
}s[1005];

ll dp[1000005];

int main() {
	ll t;
	while (cin >> t) {
		if (t < 0) {
			break;
		}
		ll sum = 0;
		for (ll i = 1; i <= t; i++) {
			cin >> s[i].x >> s[i].y;
			sum += s[i].x * s[i].y;
		}
		ll  sum1 = sum / 2;
		memset(dp, 0, sizeof(dp));
		for (ll i = 1; i <= t; i++) {
			for (ll j = 1; j <= s[i].y; j++) {
				for (ll k = sum1; k >= s[i].x; k--) {
					dp[k] = max(dp[k], dp[k - s[i].x] + s[i].x);
				}
			}
		}
		cout << sum - dp[sum1] << " " << dp[sum1] << endl;
	}
}