HDU 1171 Big Event In HDU

题目链接：​​传送门​​

Big Event in HDU

Problem Description

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).

Input

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.

Output

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.

Sample Input

2
10 1
20 1
3
10 1
20 2
30 1
-1

Sample Output

20 10
40 40

题目大意：

HDU要分家了，分成两个技术院，现在有台设备，每台设备都有一定的价值和数量，现在要使两个技术院设备的总价值尽量相同

把背包容量设成这些设备数量和价值的乘积
要尽量填起这半个背包
然后做多重背包就好了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <complex>
#include <algorithm>
#include <climits>
#include <queue>
#include <map>
#include <vector>
#include <iomanip>
#define
#define
#define

using namespace std;
int n, V, w[A], c[A], f[A]; 

int main() {
    while (cin >> n) {
        if (n < 0) return 0;
        memset(f, 0, sizeof f); V = 0;
        for (int i = 1; i <= n; i++) {
            cin >> w[i] >> c[i];
            V += w[i] * c[i];
        }
        for (int i = 1; i <= n; i++)
          for (int k = 0; k < c[i]; k++)
              for (int j = V / 2; j >=w[i]; j--)
              f[j] = max(f[j], f[j - w[i]] + w[i]);
        cout << max(f[V / 2], V - f[V / 2]) << " " << min(f[V / 2], V - f[V / 2]) << endl;
    }
}