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
題目比較簡單!
題意： n件物品，價(jià)值有正有負(fù)，最多買m件，求最多能賺多少錢。
直接排序，只取負(fù)值，最多取m個(gè)
代碼：
#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;}