Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 22611    Accepted Submission(s): 7942

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

题意：给出每个物体的价值和物体的数量，如何分使得A,B所得价值最接近并且A的价值不能小于B

<span style="font-size:18px;">#include<stdio.h>
#include<iostream>
using namespace std;
int dp[500000];
int v[10000];
int max(int x,int y)
{
	return (x>y?x:y);
}
int main ()
{
	int t,a,b,i,j;
	int sum,l;
	while(cin>>t)
	{
		if(t<0) break;
		sum=0;
		l=0;
		memset(dp,0,sizeof(dp));
		memset(v,0,sizeof(v));
		for(i=1;i<=t;i++)
		{
			cin>>a>>b;
			while(b--)   // 把第i种物品 换成 b件 01背包中的物品
			{
				v[l++]=a;
				sum+=a;
			}
		}

		for(i=0;i<l;i++)    // 01背包 (l件物品)
			for(j=sum/2;j>=v[i];j--)   // "背包容量"为总价值的一半
				dp[j]=max(dp[j],dp[j-v[i]]+v[i]);

			cout<<sum-dp[sum/2]<<' '<<dp[sum/2]<<endl;
	}
	return 0;
}</span>

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