題目來源： http://codeforces.com/problemset/problem/158/A

"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 a 1, a 2, ..., a n (0 ≤ a i ≤ 100), where a i 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: a i ≥ a i + 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
outputCopy
6
inputCopy
4 2
0 0 0 0
outputCopy
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.

題目大意:

第一行輸入兩個整數n，k。第二行輸入n個整數，且這些整數是不上升序列。問這些序列中，能大於或等於第k個數的個數是多少？

題目分析：

直接的模擬題，用陣列儲存n個數。列舉這n個數，符合條件的，個數就增加1。

參考程式碼：

#include <bits/stdc++.h>
using namespace std;
int main(){
	int n, k, a[100];
	cin >> n >> k;
	int ans = 0;
	for(int i=1; i<=n; i++)
		cin >> a[i];
	for(int i=1; i<=n; i++){
		if(a[i]>=a[k] && a[i]>0)
			ans ++;
	}
	cout << ans;
	return 0;
}