H

H - Big Event in HDU

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).

InputInput 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.
OutputFor 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：给出每个物体的价值和物体的数量，如何分使得A,B所得价值最接近 并且A的价值不能小于B

思路：将总和平分后，就是一道01背包题了  (之前的思路是将物品的数目折半 一直没想出来 后来在同学的提醒下把总价值折半 就很好做了) 值得注意的是 因为A大于B 所以先输入总和减去背包容量(因为背包不一定装满)

#include <bits/stdc++.h>

using namespace std;

int v[10001];
int main()
{
    int n;
    while(cin>>n,n>0)
    {
        memset(v,0,sizeof(v));
        int sum=0,ans=0,maxsum=-100000,k=1,st,en;
        for(int i=0;i<n;i++)
        {
            cin>>v[i];
            if(v[i]<0)
                sum++;
        }
        for(int i=0;i<n;i++)
        {
            ans+=v[i];
            if(maxsum<ans)
            {
                maxsum=ans;
                en=i;
                st=k;
            }
            if(ans<0)
            {
                ans=0;
                k=i+1;
            }
        }
        if(sum==n)
        {
            cout<<"0"<<" "<<v[0]<<" "<<v[n-1]<<endl;
        }
        else if(n==1)
        {
            cout<<v[0]<<" "<<v[0]<<" "<<v[0]<<endl;
        }
        else
        {
            cout<<maxsum<<" "<<v[st]<<" "<<v[en]<<endl;
        }
    }
    return 0;
}