Big Event in HDU

//转 背包做

题意：求把总价值分为两个数，使这两个数接近相等，而且这两个数必须由所有设备中的其中几种设备价值构成，并先输出比较大的数，再输出另一个数。

解题思路：DP算法，背包问题，求法是先求出总价值sum，再用dp[]求sum/2最多能放多少价值！即可以求出其中一个数了，另一个就是sum-dp[sum/2]了……
状态：f[j]:表示软件学院取得容量为j时能获得的最大值，j<=sum/2; 状态转移：f[j]=max{f[j], f[j-v[i]]+v[i]}

注意数组的大小（0<n<=50，0<v<=50，0<m<=100）

 

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

 

#include<stdio.h>

#include<string.h>



int s[100005];

int dp[222222];

int max(int a,int b)

{

return a>b?a:b;

}



int main()

{

int n,v,m;

while(~scanf("%d",&n)&&n>0)

{

int k=0;

int sum=0;

memset(s,0,sizeof(s));

memset(dp,0,sizeof(dp));

for(int i=0;i<n;i++)

{

scanf("%d %d",&v,&m);

while(m--)

{

s[k++]=v;//将价值存入数组  

sum+=v;

}

}

for(int i=0;i<k;i++)

for(int j=sum/2;j>=s[i];j--)

{

dp[j]=max(dp[j],dp[j-s[i]]+s[i]);

}

printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);

}

return 0;

}