Big Event in HDU（多重背包）

Big Event in HDU

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

问题链接：https://vjudge.net/contest/280387#problem/B

题意：两个学院平均分设备，尽量让两者差值最小，并且保证A>=B.

解题思路：尽量使AB两者价值接近,并且A分得的价值不少于B,那么问题就可以转化成B分得的价值不超过总价值的一半(即为背包的最大容量),挑选若干件物品装入背包后,使得总价值最大,即为B的价值,而A的价值就是sum-dp[sum/2].

AC代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int P[5005];
int dp[255555]; 
int main()
{    
	int n,a, b;
	int i, v, j, sum;
	while(~scanf_s("%d",&n),n>0) 
	{        
	memset(P,0,sizeof(P));     
	memset(dp,0,sizeof(dp));     
	j = 0;        
	sum = 0;       
	for(i = 0; i<n; i++)        
	{            
		scanf_s("%d%d",&a,&b);            
		while(b--)            
		{                
			P[j++] = a;              
			sum+=a;            
		}       
	}        
	for(i = 0; i<j; i++)           
		for(v = sum/2; v>=P[i]; v--)             
			dp[v] = max(dp[v],dp[v-P[i]]+P[i]);        
	printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);    
	}     
	return 0;
}