杭电1171Big Event in HDU

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29933    Accepted Submission(s): 10501

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

 

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 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

 

这道题最简单的方法应该是母函数的，这里用了01背包。

附代码：

#include
#include
#include
using namespace std;
long ans[110000],weight[110000],weath[110000],i,j,k,l,m,n,sum,a,b;
int main()
{
	while(scanf("%ld",&k)&&k>=0)
	{
		memset(ans,0,sizeof(ans));
		int flag=1;
		sum=0;
		m=0;
		for(i=1;i<=k;i++)
		{
			scanf("%ld%ld",&a,&b);
			sum+=a*b;//先求出总的 
			m+=b;//和总的个数 
			for(j=1;j<=b;j++)
			weath[flag++]=a;
		}
		for(i=1;i=weath[i];j--)
		ans[j]=max(ans[j],ans[j-weath[i]]+weath[i]);//01背包 
		m=max(sum-ans[sum/2],ans[sum/2]);//m一定不比sum/2大 
		printf("%ld %ld\n",m,sum-m);
	}
	return 0; 
}