158A - Next Round

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本文介绍了一个简单的算法，用于计算竞赛中晋级下一轮的参赛者数量。输入包括参赛者的总人数、晋级名额及按非递增顺序排列的成绩列表。文章通过示例解释了如何筛选出符合条件的参赛者。

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A. Next Round
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

“Contestant who earns a score equal to or greater than the k-th place finisher’s score will advance to the next round, as long as the contestant earns a positive score…” — an excerpt from contest rules.

A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round.
Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space.

The second line contains n space-separated integers a1, a2, …, an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1).
Output

Output the number of participants who advance to the next round.
Examples
Input

8 5
10 9 8 7 7 7 5 5

Output

Input

4 2
0 0 0 0

Output

Note

In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.

In the second example nobody got a positive score.

题意：给你n个人的成绩，问你成绩大于等于第k个人的分数且成绩为正的人的数目。
code:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
using namespace std;

const int maxn=50;

int main()
{
    int n,k;
    int a[maxn+10];
    int ans=0;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        ans=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",a+i);
        }
        for(int i=0;i<n;i++)
        {
            if(a[i]>=a[k-1]&&a[i]>0)
                ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}