CF——Next Round

最新推荐文章于 2022-06-06 13:30:27 发布

原创最新推荐文章于 2022-06-06 13:30:27 发布 · 819 阅读

0 ·

CC 4.0 BY-SA版权

Codeforces 专栏收录该内容

13 篇文章

订阅专栏

本文解析了比赛晋级规则，包括如何根据分数决定晋级人数，输入输出规范及实例演示。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

"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, ..., 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.

Sample Input

Input

8 5
10 9 8 7 7 7 5 5

Output

Input

4 2
0 0 0 0

Output

Hint

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.

注意数组是降序排列的，求能够顺利晋级的人数

#include <stdio.h>

int main()
{
    int n,i,j,k,sc;
    int score[200];
    scanf("%d%d",&n,&k);
    for(i = 1;i <= n;i++)
    {
        scanf("%d",&score[i]);
    }
    j = k;
    if(score[k] > 0)
    {
        for(i = k+1;i <= n;i++)
        {
            if(score[i] < score[k])
                break;
            j++;
        }
    }
    else
    {
        j--;
        for(i = k-1;i >= 1;i--)
        {
            if(score[i] > 0)
                break;
            j--;
        }
    }
    printf("%d\n",j);
    return 0;
}