问题： 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.
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.
思路：
1.我们记所有大于等于第k位的分数的人中序号最大的那个人序号为pos1；
2.我们记所有分数大于等于0的人中序号最大的那个人序号为pos2；
3.min(pos1,pos2)即为输出结果；
代码：
#include<iostream>
using namespace std;
int main(){
int n,k,s;
scanf("%d%d",&n,&k);
int pos1=k,pos2=n;
for(int i=1;i<=n;i++){
int score;
scanf("%d",&score);
if(i==k) s=score;
if(i>k&&score==s) pos1++;
if(pos2==n&&!score) pos2=i-1;
}
printf("%d",min(pos1,pos2));
return 0;
}