Next Round

文章目录

一、Next Round

总结

一、Next Round

本题链接：Next Round

题目：

A. Next Round

time limit per test3 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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.

本博客给出本题截图：

题意：比第k个数大的且不为0的个数

AC代码

#include <iostream>
using namespace std;
const int N = 110;
int s[N];
int main()
{
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i ++ ) cin >> s[i];
  int res = 0;
  for (int i = 1; i <= n; i ++ )
    if (s[i] && s[i] >= s[k])
      res ++;
  cout << res << endl;
  return 0;
}

总结

equal 相等的

sequence 顺序

水题，不解释