H - Big Event in HDU <HDU 1171>

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;
}

H - Big Event in HDU <HDU 1171>

原文：https://www.cnblogs.com/xuxua/p/9275474.html