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.

 

 

 

 

很简单的题目~~~~~

#include<stdio.h>
int mm[120];
int main()
{
    
    
    int n,k;
    
    
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        for(int i=1;i<=n;i++)
          scanf("%d",&mm[i]);
        int res=k;
        if(mm[k]>0)
        {
            for(int i=k+1;i<=n;i++)
            {
                if(mm[i]<mm[k])break;
                res++;
            }    
        }  
        else
        {
            res--;
            for(int i=k-1;i>=1;i--)
            {
                if(mm[i]>0) break;
                res--;
            }    
        } 
        
        printf("%d\n",res);     
    }    
    
    
    return 0;
}

 


文章来源:http://www.cnblogs.com/kuangbin/archive/2012/03/05/2380621.html