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Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52270 Accepted Submission(s): 17842

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 多重背包題，兩邊價值盡可能相近只需要一方的價值逼近總價值的一半即可，問題轉化為容量為sum/2的多重背包，但多重背包算法會超時，優化為01背包即可。 AC代碼如下：

#include <iostream>
#include <string.h>
#include <algorithm>

using namespace std;

int a[5005];
int dp[125025];

bool cmp(int a,int b)
{
    return a > b;
}
int main(void)
{
    int n;
    while (scanf_s("%d", &n))
    {
        if (n < 0)
            break;
        memset(dp, 0, sizeof(dp));
        memset(a, 0, sizeof(a));
        int sum = 0;
        int cnt = 0;
        while (n--)
        {
            int x, y;
            scanf_s("%d%d", &x, &y);
            sum += x * y;
            while (y--)
            {
                a[cnt++] = x;
            }
        }
        sort(a, a + cnt,cmp);
        for (int i = 0; i < cnt; i++)
        {
            for (int j = sum / 2; j >= a[i]; j--)
                dp[j] = max(dp[j], dp[j - a[i]] + a[i]);
        }
        int m = dp[sum / 2];
        int n = sum - m;
        if (m > n)
            cout << m << ‘ ‘ << n << endl;
        else
            cout << n << ‘ ‘ << m << endl;
    }
    return 0;
}

　　

HDUOJ 1171