国内最全IT社区平台 联系我们 | 收藏本站
华晨云阿里云优惠2
您当前位置:首页 > php开源 > php教程 > 158A - Next Round

158A - Next Round

来源:程序员人生   发布时间:2014-12-13 09:19:02 阅读次数:3960次
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.

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,码农辛苦
如果您觉得本网站对您的学习有所帮助,可以手机扫描二维码进行捐赠
程序员人生
------分隔线----------------------------
分享到:
------分隔线----------------------------
关闭
程序员人生