Big Event in HDU
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
题意描述：
给你n件物品，每件物品给你了价值和数量，让你价值尽量公平的把所有物品放到两个包里，求每个包各放多少价值的物品，我在做的时候是把所有物品的总价值求出来，然后往一个包里放，直到放到一半或接近一半时停止，剩下的放到另一个包里
程序代码：
#include<stdio.h>
#include<algorithm>
using namespace std;
#include<string.h>
int v[300010],w[300010],dp[300010];
int main()
{
int T,n,m,i,j,k,sum;
while(scanf("%d",&n)!=EOF)
{
if(n<0)
break;
sum=0;
memset(dp,0,sizeof(dp));
for(i=0;i<n;i++)
{
scanf("%d%d",&v[i],&w[i]);
sum+=v[i]*w[i];
}
for(i=0;i<n;i++)
for(j=0;j<w[i];j++)
for(k=sum/2;k>=v[i];k--)
dp[k]=max(dp[k],dp[k-v[i]]+v[i]);
printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);
}
return 0;
}