Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18273    Accepted Submission(s): 6403

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

A easy one ,but the question producer is so boring that he make such fucking test cases.The input is terminated with a negative integer not minus one.Beside when N equals to zero the case is also not to be processed too.

#include<stdio.h>

#include<string.h>

bool f[300000];

int v[125],m[125];

int main()

{

	int N;

	while (scanf("%d",&N)!=EOF)

	{

		if (N<0) return 0;

		if (N==0) continue;

		int sum=0,i,j,k,A;

		for (i=1;i<=N;i++)

		{

			scanf("%d%d",&v[i],&m[i]);

			sum+=v[i]*m[i];

		}

		memset(f,0,sizeof(f));

		f[0]=true;

		for (i=1;i<=N;i++)

			for (j=1;j<=m[i];j++)

				for (k=sum;k>=0;k--)

					if (f[k]) f[k+v[i]]=true;

		for (i=(sum+1)/2;i<=sum;i++)

			if (f[i])

			{

				A=i;

				break;

			}

		printf("%d %d\n",A,sum-A);

	}

	return 0;

}