[CF从零单排#3] CF158A - Next Round

Posted 2020-12-18 gdgzliu

题目来源： http://codeforces.com/problemset/problem/158/A

"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 a 1,?a 2,?...,?a n (0?≤?a i?≤?100), where a i 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).

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.

题目大意:

第一行输入两个整数n，k。第二行输入n个整数，且这些整数是不上升序列。问这些序列中，能大于或等于第k个数的个数是多少？

题目分析：

直接的模拟题，用数组保存n个数。枚举这n个数，符合条件的，个数就增加1。

参考代码：

#include <bits/stdc++.h>
using namespace std;
int main(){
	int n, k, a[100];
	cin >> n >> k;
	int ans = 0;
	for(int i=1; i<=n; i++)
		cin >> a[i];
	for(int i=1; i<=n; i++){
		if(a[i]>=a[k] && a[i]>0)
			ans ++;
	}
	cout << ans;
	return 0;
}