Hust oj 1369 Buy Computers（水题）

最新推荐文章于 2021-02-27 16:50:52 发布

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最新推荐文章于 2021-02-27 16:50:52 发布

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分类专栏：水题文章标签：哈理工oj1369

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Buy Computers

Time Limit: 1000 MS

Memory Limit: 65536 K

Total Submit: 150(78 users)

Total Accepted: 79(75 users)

Rating:

Special Judge: No

Description

Leyni goes to a second-hand market for old computers. There are n computers at the sale and computer i costs c_i Yuan. Some computers with a negative price mean that their owners will pay Leyni if he buys them. Leyni can carry at most m computers, and he won’t go to the market for a second time.

Leyni wonders the maximum sum of money that he can earn.

Input

There are multiple test cases. The first line of input is an integer T indicating the number of test cases. Then T test cases follow.

For each test case:

Line 1. This line contains two space-separated integers n and m (1 ≤ m ≤ n ≤ 100) indicating the amount of computers at the sale and the amount of computers that Leyni can carry.

Line 2. This line contains n space-separated integers c_i (-1000 ≤ c_i ≤ 1000) indicating the prices of the computers.

Output

For each test case:

Line 1. Output the maximum sum of money that Leyni can earn.

Sample Input

5 3

-5 3 2 1 -4

Sample Output

Source

哈理工2012春季校赛 - 网络预选赛

Author

齐达拉图@HRBUST

问能挣多少钱，那就是问小于m的最大负数和是多少。。大水题

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn = 105;
int sum;
int n,m;
int t;
int a[maxn];

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        sum = 0;
        int num = 0;
        sort(a,a+n);
        for(int i=0;i<n;i++)
        {
            if(a[i] < 0 && num < m)
            {
                num++;
                sum += a[i];
            }
        }
        printf("%d\n",-sum);
    }
    return 0;
}