158A

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

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.

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;}