AtCoder Beginner Contest 171 B

B - Mix Juice

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Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $200200$ points

Problem Statement

A shop sells $\mathrm{NN}$ kinds of fruits, Fruit $\mathrm{1,\dots ,N1,\dots ,N}$, at prices of ${\mathrm{p1,\dots ,pNp1,\dots ,pN}}^{}$ yen per item, respectively. (Yen is the currency of Japan.)

Here, we will choose $\mathrm{KK}$ kinds of fruits and buy one of each chosen kind. Find the minimum possible total price of those fruits.

Constraints

• $\mathrm{1\le K\le N\le 10001\le K\le N\le 1000}$
• $\mathrm{1\le pi\le 10001\le pi\le 1000}$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$\mathrm{NN}$ $\mathrm{KK}$
${\mathrm{p1p1}}^{}$ ${\mathrm{p2p2}}^{}$ $\dots \dots$ ${\mathrm{pNpN}}^{}$

Output

Print an integer representing the minimum possible total price of fruits.

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Sample Input 1 Copy

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5 3
50 100 80 120 80

Sample Output 1 Copy

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210

This shop sells Fruit $11$, $22$, $33$, $44$, and $55$ for $5050$ yen, $100100$ yen, $8080$ yen, $120120$ yen, and $8080$ yen, respectively.

The minimum total price for three kinds of fruits is $\mathrm{50+80+80=21050+80+80=210}$ yen when choosing Fruit $11$, $33$, and $55$.

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Sample Input 2 Copy

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1 1
1000

Sample Output 2 Copy

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1000
题意：输入一个n，然后输入n个数，找出前k小的数的和，并输出
思路:此题解法多种，可以是用优先队列来做也可以对数组排序来做。
优先队列：

#include<cstdio>
#include<queue>
using namespace std;
int main(void)
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        priority_queue<int,vector<int>,greater<int> >p;//小顶堆
        for(int i=0;i<n;++i)
        {
            int t;
            scanf("%d",&t);
            p.push(t);
        }
        int sum=0;
        for(int i=0;i<k;++i)
        sum+=p.top(),p.pop();
        printf("%d
",sum);
        while(!p.empty())
        p.pop();
    }
    return 0;
}

排序：

#include<cstdio>
#include<algorithm>
#define maxn 1005
using namespace std;
bool cmp(int x,int y)
{
    return x<y;
}
int main(void)
{
    int n,k;
    int a[maxn];
    while(~scanf("%d%d",&n,&k))
    {
        for(int i=0;i<n;++i)
            scanf("%d",&a[i]);
            sort(a,a+n,cmp);
        int sum=0;
        for(int i=0;i<k;++i)
        sum+=a[i];
        printf("%d
",sum);
    }
    return 0;
}

关于sort的用法可以参考这个博客：https://www.cnblogs.com/YHH520/p/12253671.html