Codeforces VK Cup 2012 Qualification Round 1 / 158A Next Round(模拟)

本文解析了Codeforces平台上的竞赛晋级规则,并通过一个具体的题目示例解释如何根据参赛者的分数来确定晋级人数。示例中包含输入输出样例及特殊判断情况。

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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.

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.


注意特判。


完整代码:

/*30ms,0KB*/

#include<cstdio>

int main()
{
	int n, k, a, i, now;
	scanf("%d%d", &n, &k);
	for (i = 1; i <= n; ++i)
	{
		scanf("%d", &a);
		if (a == 0)
		{
			printf("%d", i - 1);
			return 0;
		}
		if (i == k)
		{
			now = a;
		}
		else if (i > k)
		{
			if (a != now)
			{
				printf("%d", i - 1);
				return 0;
			}
		}
	}
	printf("%d", i - 1);
	return 0;
}



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