CodeForces 158A Next Round

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 andk (1 ≤ k ≤ n ≤ 50) separated by a single space.

The second line contains n space-separated integersa1, a2, ..., an
(0 ≤ ai ≤ 100), whereai is the score earned by the participant who got thei-th
place. The given sequence is non-increasing (that is, for alli 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.

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

bool complare(int a,int b){
return a>b;
}

int main(){
int n,k,a[110],i,sum,hello;
while(scanf("%d %d",&n,&k)!=EOF){
sum=0;
hello=-1;
for(i=0;i<n;i++){
scanf("%d",&a[i]);
}
sort(a,a+n,complare);
hello=a[k-1];
for(i=0;i<n;i++){
if(hello==0){
if(a[i]>hello){
sum++;
}
}
else if(a[i]>=hello){
sum++;
}
}
printf("%d\n",sum);
memset(a,0,sizeof(a));
}
return 0;
}