CodeForces 34B Sale

B - SaleTime Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64uSubmit Status Practice CodeForces 34BDescriptionOnce 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.InputThe 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.OutputOutput the only number — the maximum sum of money that Bob can earn, given that he can carry at most m TV sets.Sample InputInput

5 3
-6 0 35 -2 4

Output

8

Input

4 2
7 0 0 -7

Output

7

很水的一个题，看数据猜题意过的，先排序然后选前m小的数，如果前m个数有大

于0的数则提前终止，最后把前m小的数字相加在乘以-1，得到一个正数输出就可以了

AC代码

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<iostream>#include<algorithm>using namespace std;int main(){    int n,m;    int a[10001];    while(scanf("%d%d",&n,&m)!=EOF)    {        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        sort(a,a+n);        int sum = 0;        for(int i=0;i<m;i++)        {            if(a[i]<0)            {                sum = sum + (-1*a[i]);            }            else            {                break;            }        }        printf("%d\n",sum);    }    return 0;}