158A_Next_Round.cpp

/*                                        A. Next Round
time limit per test3 seconds
memory limit per test256 megabytes
"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
InputCopy
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.*/

#include <bits/stdc++.h>
using namespace std;

int main() {
    int n, k, c = 0;
    cin >> n >> k;

    vector<int> arr(n);
    for (int i = 0; i < n; ++i) {
        cin >> arr[i];
    }

    int x = arr[k - 1]; 

    for (int i = 0; i < n; ++i) {
        if (arr[i] >= x && arr[i] > 0) { 
            c++;
        }
    }

    cout << c << endl;
    return 0;
}