"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
Copy
8 5 10 9 8 7 7 7 5 5
output
Copy
6
input
Copy
4 2 0 0 0 0
output
Copy
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.
题意：长度为n的有序降序序列，成绩比第k名高或者相等的且不为0的人数
#include<iostream>
#include<string>
#include<cstring>
usingnamespace std;
typedef longlong ll;
int a[1100];
int main()
{
int n,k;
cin>>n>>k;
for(int i=1;i<=n;i++) cin>>a[i];
int cnt=k;
for(int i=k+1;i<=n;i++)
{
if(a[i]>=a[k]) cnt++;
}
int tmp=cnt;
for(int i=1;i<=tmp;i++) if(a[i]==0) cnt--;
cout<<cnt;
return0;
}
