hduoj1171!【数学】

/*Big Event in HDU
Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23323    Accepted Submission(s): 8217


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
 
Author
lcy
*/
#include<stdio.h>
#include<stdlib.h>
int cmp(const void *a, const void *b)
{
	return *(int *)a - *(int *)b;
}
int num1[100000];
int main()
{
    int n, i, j, m, sum,  num;
    while(1)
    {
		scanf("%d", &n);
		if(n < 0)
		break;
		sum = 0;
		j = 0;
		for(i = 0; i  < n; i++)
		{
			scanf("%d %d", &m, &num);
            		sum += m * num;
            		while(num--)
           		  num1[j++] = m;        //将每个数都保存进数组
		}
		qsort(num1, j, sizeof(num1[0]), cmp);
		num = 0;
		for(i = j-1;  i >= 0; i--)
		{
			if(num + num1[i] > sum / 2)// 将能加并且加了后小于总和的一半的数全部加进去
			continue;
			num += num1[i];
		}
		printf("%d %d\n", sum - num, num);
	}
	return 0;
}

题意：给定几个数以及每个数所对应的个数，尽量平分成2部分。

本题可以用母函数做，不过感觉很麻烦，这个易懂省事。