Codeforces 34 B. Sale_b. sale codeforces-优快云博客

本文链接：https://blog.youkuaiyun.com/baodream/article/details/80376375

本文解析了CodeForces上的一道经典问题，通过购买负价格电视机获取最大收益的算法思路。介绍了输入输出格式及样例，给出了简洁高效的C++实现代码。

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题目链接：http://codeforces.com/contest/34/problem/B

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.

  Examples
 
    input
    
     Copy
    
5 3
-6 0 35 -2 4

    output
    
     Copy
    
8

    input
    
     Copy
    
4 2
7 0 0 -7

    output
    
     Copy
    
7

题目大意：

给定n台电视机价格（有负数，为什么？我也不知道为什么有负数），Bob最多可以买m台，问Bob可以获得的最大价值。

题目思路：

把价格取反，排序之后把所有正数加起来即可，注意，最多m个。

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const double DINF = 0xffffffffffff;
const int mod = 1e9+7;
const int N = 1e6+5;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    int n,m;
    int a[105];
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        a[i] = -a[i];
    }
    sort(a+1,a+1+n);
    int ans=0;
    for(int i=n;i>=1&&i>n-m;i--)
    {
        if(a[i]>0)
            ans+=a[i];
        else
            break;
    }
    cout<<ans<<endl;

    return 0;
}