HDU 1171 Big Event in HDU

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36300    Accepted Submission(s): 12611

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

小编刚学0-1背包，在一个大神博客里看了0-1背包分类，开始刷，其实对于这个题我感觉不像是0-1背包，但确实解决了。

题意大概就是给你设备的价值和数量，让你均分。我是把总价值的一半看成B的背包容量（因为A>=B），价值就是价值哈，建立状态转移方程，对了，存储的时候有点小小技巧，建议看代码哈。

#include <stdio.h>
#include <string.h>

int max(int x,int y)
{
	return x>y?x:y;
}
int main()
{
	int i,j,n,v[100005],dp[100005];
	int a,b,sum,cnt;
	
	while(~scanf("%d",&n),n>=0)
	{
		sum=cnt=0;
		j=1;
		while(n--)
		{
			scanf("%d%d",&a,&b);
			while(b--)
			{
				v[j++]=a;
				sum+=a;
				cnt++;	
			}
			
		}
		memset(dp,0,sizeof(dp));
		for(i=1;i<=cnt;i++)
		{
			for(j=sum/2;j>=v[i];j--)
			{
				dp[j]=max(dp[j],dp[j-v[i]]+v[i]);
			}
		}
		
		printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);
	}
	return 0;
}