== tip days 很好 enter lines ans sta ati

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

題意：給出每個物體的價值和物體的數量，如何分使得A,B所得價值最接近 並且A的價值不能小於B

思路：將總和平分後，就是一道01背包題了 (之前的思路是將物品的數目折半 一直沒想出來 後來在同學的提醒下把總價值折半 就很好做了) 值得註意的是 因為A大於B 所以先輸入總和減去背包容量(因為背包不一定裝滿)

#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;
}

H - Big Event in HDU <HDU 1171>