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题目

A. Next Round
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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
inputCopy
8 5
10 9 8 7 7 7 5 5
outputCopy
6
inputCopy
4 2
0 0 0 0
outputCopy
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.

代码

#include<stdio.h>
int main()
{
int n,k,j,i,a[101],num=0;
scanf("%d %d",&n,&k);
for(i=0;i<n;i++){
scanf("%d",&a[i]);
}
for(i=0;i<n;i++){
if(a[i]>=a[k-1]&&a[i]!=0){
num++;
}else{
break;
}
}
printf("%d",num);
return 0;
}

总结

英语不好，刚开始没理解题目意思，还不太习惯…
n是人数，k是排名，分数比第k名的高的和相等的 且 分数不为零的人可以晋级。这就要从头开始判断分数，满足以上两个条件。
刚开始我没考虑到分数都为零的情况，像第二个例子，直接从第k名往后判断…
需要注意的是，如果数组是从a[0]开始放数据，那么第k名在数组里应该是a[k-1]，而不是a[k]。

最后

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