codeforces-34B-B. Sale( C++ && greedy && sortings )

Once Bob got to a sale of old TV sets. There were n TV sets at that sale. TV set with index i costs ai bellars. Some TV sets have a negative PRice — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most m TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.

The first line contains two space-separated integers n and m (1 ≤ m ≤ n ≤ 100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains n space-separated integers ai ( - 1000 ≤ ai ≤ 1000) — prices of the TV sets.

Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most m TV sets.

5 3-6 0 35 -2 4output
8input
4 27 0 0 -7output
7
题目比较简单!
题意： n件物品，价值有正有负，最多买m件，求最多能赚多少钱。
直接排序，只取负值，最多取m个
代码：
#include<iostream>#include<cstdio>#include<algorithm>using namespace std;int main(){    int n,m,sum=0;    scanf("%d%d",&n,&m);    int a[n]={0};    for(int i=0;i<n;i++){        scanf("%d",&a[i]);    }    sort(a,a+n);    for(int i=0;i<m;i++){        if(a[i]<0){            sum+=a[i];        }    }    printf("%d/n",-sum);    return 0;}