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A. Next Round
点击打开题目
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
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8 5 10 9 8 7 7 7 5 5
output
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6
input
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2
4 2 0 0 0 0
output
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1
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.

第一行输入一个n和k,分别表示n个人,和排名为k的位置,第二行输入n个从大到小的数排名从1到n的。

代码:

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#include <iostream> #include <math.h> using namespace std; int main() { int a[100],n,k,ans,i; while(cin>>n>>k) { ans=0; for(i=0;i<n;i++) { cin>>a[i]; } for(i=0;i<n;i++) { if(a[i]>=a[k-1]&&a[i]!=0)//注意下标与排名是差一的 ans++; } cout<<ans<<endl; } return 0; }


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