A. Next Round
time limit per test: 3 second
memory limit per test: 256 megabytes
"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.
Example 1
| Input |
|---|
| 8 5 |
| 10 9 8 7 7 7 5 5 |
| Output |
|---|
| 6 |
Example 2
| 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.
Solution
Try to solve it on your own before looking at the solution.
Python3 / Python2 / PyPy / ...
# Leer n (participantes) y k (posición límite)
n, k = map(int, input().split())
# Leer las puntuaciones como una lista de enteros
scores = list(map(int, input().split()))
# Obtener la puntuación del k-ésimo participante (índice k-1 porque las listas empiezan en 0)
kth_score = scores[k - 1]
advancers = 0
# Contar cuántos avanzan
for score in scores:
# Deben tener un puntaje positivo y al menos igual al k-ésimo lugar
if score > 0 and score >= kth_score:
advancers += 1
else:
# Como la lista ya está ordenada de forma no creciente,
# si uno no cumple, los demás tampoco lo harán
break
print(advancers)You are welcome to share your solution in another programming language