Python:实现使用矩阵求幂的第 n 个斐波那契算法
def multiply(matrix_a, matrix_b):
matrix_c = []
n = len(matrix_a)
for i in range(n):
list_1 = []
for j in range(n):
val = 0
for k in range(n):
val = val + matrix_a[i][k] * matrix_b[k][j]
list_1.append(val)
matrix_c.append(list_1)
return matrix_c
def identity(n):
return [[int(row == column) for column in range(n)] for row in range(n)]
def nth_fibonacci_matrix(n):
"""
>>> nth_fibonacci_matrix(100)
354224848179261915075
>>> nth_fibonacci_matrix(-100)
-100
"""
if n <= 1:
return n
res_matrix = identity(2)
fibonacci_matrix = [[1, 1], [1, 0]]
n = n - 1
while n > 0:
if n % 2 == 1:
res_matrix = multiply(res_matrix, fibonacci_matrix)
fibonacci_matrix = multiply(fibonacci_matrix, fibonacci_matrix)
n = int(n / 2)
return res_matrix[0][0]
def nth_fibonacci_bruteforce(n):
"""
>>> nth_fibonacci_bruteforce(100)
354224848179261915075
>>> nth_fibonacci_bruteforce(-100)
-100
"""
if n <= 1:
return n
fib0 = 0
fib1 = 1
for i in range(2, n + 1):
fib0, fib1 = fib1, fib0 + fib1
return fib1
def main():
for ordinal in "0th 1st 2nd 3rd 10th 100th 1000th".split():
n = int("".join(c for c in ordinal if c in "0123456789")) # 1000th --> 1000
print(
f"{ordinal} fibonacci number using matrix exponentiation is "
f"{nth_fibonacci_matrix(n)} and using bruteforce is "
f"{nth_fibonacci_bruteforce(n)}\n"
)
# from timeit import timeit
# print(timeit("nth_fibonacci_matrix(1000000)",
# "from main import nth_fibonacci_matrix", number=5))
# print(timeit("nth_fibonacci_bruteforce(1000000)",
# "from main import nth_fibonacci_bruteforce", number=5))
# 2.3342058970001744
# 57.256506615000035
if __name__ == "__main__":
import doctest
doctest.testmod()
main()
