Codeforces 34B-Sale（暴力）

B. Sale

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

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 TV sets.

Sample test(s)

input

5 3-6 0 35 -2 4

output

8

input

4 27 0 0 -7

output

7

也是水题。。

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

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <string>#include <cctype>#include <vector>#include <cstdio>#include <cmath>#include <deque>#include <stack>#include <map>#include <set>#define ll long long#define maxn 116#define pp pair<int,int>#define INF 0x3f3f3f3f#define max(x,y) ( ((x) > (y)) ? (x) : (y) )#define min(x,y) ( ((x) > (y)) ? (y) : (x) )using namespace std;int n,m,a[102];int main(){    while(scanf("%d%d",&n,&m)!=EOF){for(int i=0;i<n;i++)scanf("%d",a+i);;sort(a,a+n);int ans=0,cnt=0;for(int i=0;i<n;i++)if(a[i]<0&&cnt<m){ans+=a[i];cnt++;}elsebreak;printf("%d\n",-ans);}return 0;}