Q>
You are given an array with positive numbers and a number N.
(양수와 수 N의 배열이 주어집니다.)
You should find the N-th power of the element in the array with the index N.
(배열의 N 번째 지수는 N입니다.)
If N is outside of the array, then return -1.
(N이 배열 외부에 있으면 -1을 반환합니다.)
Don't forget that the first element has the index 0.
(첫 번째 요소의 인덱스가 0임을 잊지 마십시오.)
Let's look at a few examples:
(몇 가지 예를 살펴 보겠습니다. )
- array = [1, 2, 3, 4] and N = 2, then the result is 32 == 9;
- array = [1, 2, 3] and N = 3, but N is outside of the array, so the result is -1.
Input: Two arguments. An array as a list of integers and a number as a integer.
(두 개의 인수. 정수의리스트로서의 배열. 정수로서의 숫자.)
Output: The result as an integer.
(결과는 정수입니다.)
Example:
index_power([1, 2, 3, 4], 2) == 9
index_power([1, 3, 10, 100], 3) == 1000000
index_power([0, 1], 0) == 1
index_power([1, 2], 3) == -1
How it is used:
This mission teaches you how to use basic arrays and indexes when combined with simple mathematics.
(이 임무는 간단한 수학과 결합 할 때 기본 배열과 색인을 사용하는 방법을 알려줍니다.)
Precondition:
0 < len(array) ≤ 10
0 ≤ N
all(0 ≤ x ≤ 100 for x in array)
A>
def index_power(array: list, n: int) -> int:
# n이 리스트의 값보다 작은가?
if n < len(array):
# 작으면 n번 인덱스를 가져와서 n을 제곱
return array[n] ** n
# 그 이외는 -1
else:
return -1
if __name__ == '__main__':
print('Example:')
print(index_power([1, 2, 3, 4], 2))
# These "asserts" using only for self-checking and not necessary for auto-testing
assert index_power([1, 2, 3, 4], 2) == 9, "Square"
assert index_power([1, 3, 10, 100], 3) == 1000000, "Cube"
assert index_power([0, 1], 0) == 1, "Zero power"
assert index_power([1, 2], 3) == -1, "IndexError"
print("Coding complete? Click 'Check' to review your tests and earn cool rewards!")
O>
Example:
9
Coding complete? Click 'Check' to review your tests and earn cool rewards!
Process finished with exit code 0
S>
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