BOJ 2748フィボナッチ数2

https://www.acmicpc.net/problem/2748
時間1秒、メモリ128 MB
input :

n (1 <= n <= 90)

output :

n 2番目のフィボナッチ数出力.

0番目を書き込み(previde)、1番目を書き込み(current).
tempを追加して交換し、値を記録します.

import sys

n = int(sys.stdin.readline())

previous = 0
current = 1

if n == 0:
    print(0)
elif n == 1:
    print(1)
else:
    for i in range(2, n + 1):
        temp = current
        current += previous
        previous = temp
    print(current)