Codeforces——158A Next Round

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我是靠谱客的博主魁梧百合，最近开发中收集的这篇文章主要介绍Codeforces——158A Next Round，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

小声BB

最近总是状态不好，不管是范围还是特例都有思考不周之处

题干

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.

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.

分析

就是跟第k名的人比分数，除了0分外，不比他小的就计次，这里要注意写循环，不要炫技，写for循环的结束条件时不考虑范围（说的就是我）

代码

#include<iostream>
using namespace std;
int main()
{
int a[100],n,k;
cin >> n >> k;
for(int i = 0;i < n;i++)
cin >> a[i];
int std = a[k - 1];
int ct = 0;
for(int i = 0;i < n;i++)
if(a[i] >= a[k - 1] && a[i] != 0)
ct++;
cout << ct;
return 0;
}