如何在python中生成函数n次并计算数字？

Question

我想根据用户输入生成斐波那契函数 N 次。 （我认为这是一个 while 循环或 if 语句，如 if N in range）。 还有第二个用户输入定义为 Y。Y 表示重复函数的位数，我想计算生成的数字有多少位数是 Y。

以下是我的不完整代码：

N = int(input("Enter N: "))
Y = int(input("Enter Y: "))


def fibonacci(n): 
   if n <= 1:
     return n
   else:
     return fibonacci(n-2) + fibonacci(n-1)

nterms = 10

# check if the number of terms is valid
if nterms <= 0:
   print("Please enter a positive integer")
else:
   print("Fibonacci sequence:")
   for i in range(nterms):
       print(fibonacci(i))

提前致谢

Answer 1

试试这个（阅读下面代码中的 cmets）：

from math import sqrt

def fibonacci(n):
    return int(((1+sqrt(5))**n-(1-sqrt(5))**n)/(2**n*sqrt(5))) # WA algorithm

# Handle N from user input
while True:
    N = int(input("Enter N: "))
    Y = int(input("Enter Y: "))
    if N < 1 or Y < 1:
        print("Plese enter a positive integers for N and Y")
    else:
        break

count = 0 # Counter of the instances of fibonacci numbers containing Y digits
for n in range(N):
    fib = fibonacci(n) # Calls fibonacci
    print(fib) # Print Fibonacci number
    if len(str(fib)) == Y: # Number of digits in the Fibonacci number
        count += 1 # Count up if number of digits = Y

print("\nFibonacci numbers with {} digits occured {} times" .format(Y, count))

计算斐波那契数的算法来自Wolfram Alpha

编辑： 为了节省查找多个Ys 的结果的时间，您可以将统计信息保存在字典中，如下所示：

from math import sqrt

# Handle N from user input
while True:
    N = int(input("Enter N: "))
    if N < 1:
        print("Plese enter a positive integers for N and Y")
    else:
        break

size = {} # Dictionary with number of occurences of lengths of Fibonacci numbers
count = 0 # Counter of the instances of fibonacci numbers containing Y digits
for n in range(N):
    fib = int(((1+sqrt(5))**n-(1-sqrt(5))**n)/(2**n*sqrt(5))) # WA algorithm
    print(fib) # Print Fibonacci number
    s = str(len(str(fib)))
    if s not in size:
        size[s] = 1
    else:
        size[s] += 1

print()
for key, val in size.items():
    print("{}\toccured\t{} times" .format(key, val))

这将产生如下输出：

Enter N: 13
0
1
1
2
3
5
8
13
21
34
55
89
144

1   occured 7 times
2   occured 5 times
3   occured 1 times