B - Sale
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Submit Status Practice CodeForces 34B
Description
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
Input
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
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most m
Sample Input
Input
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;
}
