您當前位置：首頁 > php開源 > php教程 > 158A - Next Round

158A - Next Round

來源：程序員人生發布時間：2014-12-13 09:19:02 閱讀次數：4155次

"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.

import java.util.*; public class Main { public static void main(String[] args) { Scanner sc=new Scanner(System.in); int n=sc.nextInt(); int k=sc.nextInt(); int count=0; int[] score = new int[60]; for(int i=0;i<n;i++) score[i]=sc.nextInt(); int max=score[k⑴]; for(int i=0;i<n;i++) if(score[i]>=max&&score[i]>0) count++; System.out.println(count); } }

生活不易，碼農辛苦
如果您覺得本網站對您的學習有所幫助,可以手機掃描二維碼進行捐贈
程序員人生

------分隔線----------------------------

上一篇 從零開始iOS8編程【iOS開發常用控件】

下一篇 HDU 4810 Wall Painting(異或 +按位容斥)

分享到:

------分隔線----------------------------

為碼而活

積分：4237

15粉絲

7關注

欄目熱點

日本搞逼视频_黄色一级片免费在线观看_色99久久_性明星video另类hd_欧美77_综合在线视频

158A - Next Round