HDOJ--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): 30206    Accepted Submission(s): 10606

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

 

思路：与01背包有些不同的就是，给出的物品中，有一部分物品的件数并非是一件。所以需要我们去想办法让他们尽可能的接近01背包的状态。

ac代码：

#include
#include
#include
using namespace std;
int dp[1010*1010];
int main(){
	int N;
	int weight[1010*1010],num,V,sum,i;
	while(scanf("%d",&N)!=EOF){
		if(N<0)//注意题目中的要求，对于负整数的话结束运行。 
			break;
		sum=0;
		int cnt=0,v;
		for(i=0;i=weight[i];j--)
			dp[j]=max(dp[j],dp[j-weight[i]]+weight[i]);
		}
		printf("%d %d\n",sum-dp[V],dp[V]);	 
	}
	return 0;
}