n的階乘程序python

Formula used:

使用的公式：

Trailing 0s in N! = Count of 5s in prime factors of n!= floor(n/5) + floor(n/25) + floor(n/125) + ....

Example:

例：

Input: N = 23Output: 4Factorial of 23 is 25852016738884976640000 which has four trailing 0.Input: N = 25Output: 6Factorial of 25 is 15511210043330985984000000 which has six trailing 0.

Code:

碼：

# Define a function for finding # number of trailing zeros in N! def find_trailing_zeros(num) :sum = 0i = 1# iterating untill quotient is not zerowhile(True) :# take integer divison quotient = num // (5 ** i)if(quotient == 0) :breaksum += quotienti += 1return(sum)# Driver code if __name__ == "__main__" :# assigning a numbernum = 10# function callprint("Number of trailing zeros in factorial of",num,"is :",find_trailing_zeros(num))num = 20print("Number of trailing zeros in factorial of",num,"is :",find_trailing_zeros(num))

Output

輸出量

Number of trailing zeros in factorial of 10 is : 2 Number of trailing zeros in factorial of 20 is : 4

翻譯自: https://www.includehelp.com/python/count-number-of-trailing-zeros-in-factorial-of-number-n.aspx

n的階乘程序python

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