codeforces 158A Next Round（水题）

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我是靠谱客的博主细心太阳，最近开发中收集的这篇文章主要介绍codeforces 158A Next Round（水题），觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

"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.

Sample test(s)

input
8 5
10 9 8 7 7 7 5 5

output
6

input
4 2
0 0 0 0

output
0

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.

第一行输入一个n和k，分别表示n个人，和排名为k的位置，第二行输入n个从大到小的数排名从1到n的。

代码:

#include <iostream>
#include <math.h>
using namespace std;
int main()
{
int a[100],n,k,ans,i;
while(cin>>n>>k)
{
ans=0;
for(i=0;i<n;i++)
{
cin>>a[i];
}
for(i=0;i<n;i++)
{
if(a[i]>=a[k-1]&&a[i]!=0)//注意下标与排名是差一的
ans++;
}
cout<<ans<<endl;
}
return 0;
}