以下是斐波那契数列的递归定义:
Fn = Fn−1 + Fn−2, F1 = 1,F2 = 1.
那么其12项为:
F1 = 1
F2 = 1
F3 = 2
F4 = 3
F5 = 5
F6 = 8
F7 = 13
F8 = 21
F9 = 34
F10 = 55
F11 = 89
F12 = 144
因此第12项,F12,是第一个包含三位数字的项。
斐波那契数列中第一个包含1000位数字的项是第几项?
# 按位求取
n_digit = 1000
fn_1 = [0] * n_digit
fn_1[0] = 1
fn_2 = fn_1.copy()
n = 2
digit = 1
while 1:
d = 0
fn = [0] * n_digit
for i in range(digit):
fn[i] = fn_1[i] + fn_2[i] + d
d = fn[i] // 10
fn[i] %= 10
if d > 0:
fn[digit] = d
digit += 1
n += 1
if digit >= n_digit:
break
fn_1 = fn_2.copy()
fn_2 = fn.copy()
print(n)