158A. Next Round_牛客博客

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.

“得分等于或大于第k名选手得分的选手将进入下一轮，只要该选手得分为正……”选自竞赛规则。

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.

共有n名参与者参加了比赛（n ≥ k），你已经知道他们的分数了。计算有多少参与者将进入下一轮。

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space.

输入的第一行包含两个整数n和k（1） ≤ K ≤ N ≤ 50）被一个单独的空间隔开。

The second line contains n space-separated integers a $_{1}$ , a $_{2}$ , ..., a $_{n}$ (0 ≤ a $_{i}$ ≤ 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: a $_{i}$ ≥ a $_{i} + 1$ ).

第二行包含n个空格分隔的整数a $_{1}$ ， a $_{2}$ ， ...,a $_{n}$ （0 ≤ a $_{i}$ ≤ 100），其中a $_{i}$ 是获得第i名的参与者获得的分数。给定的序列是非递增的（即，对于从1到n-1的所有i）满足以下条件：a $_{i}$ ≥ a $_{i} + 1$ ).

Output

Output the number of participants who advance to the next round.

输出进入下一轮的参与者人数。

Examples

input1

8 5

10 9 8 7 7 7 5 5

output1

input2

4 2

0 0 0 0

output2

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.

在第一个例子中，排名第五的参与者获得7分。由于第六名的参与者也获得了7分，因此有6名获得者。

In the second example nobody got a positive score.

在第二个例子中，没有人得到正面分数。

Solution

考虑两个条件：

得分等于或大于第k名选手得分（a[i] >= a[k-1]）
选手得分为正（a[i] > 0）

Code

#include <iostream>
using namespace std;

//158A. Next Round
int main() {
    int n = 0;//n：n名参与者
    int k =0;//k：得分等于或大于第k名选手得分
    cin >> n >> k;
    int sum = 0;//sum：进入下一轮的参与者人数
    int a[n];//a[i]：获得第i名的参与者获得的分数
    for(int i = 0;i < n;i++){
        cin >> a[i];
    }
    int flag = a[k-1];
    for(int i = 0;i < n;i++) {
        if (a[i] >= flag && a[i] > 0) {
            sum++;
        }else{
            break;
        }
    }
    cout << sum << endl;
    return 0;
}