实验吧 手脑并用

Posted 2020-10-11 时光不改

手脑并用

 

对N！进行数学分析

解题链接： http://ctf5.shiyanbar.com/ppc/1.txt

 

求6789！的值末尾有几个零？

 

解题思路：

好吧，这感觉是一道送分题，求6789的阶乘

因为网上这个算法很多呀，求出这个阶乘复制一下后面的0，然后求和就可以了。

 

#!/usr/bin/python3
 
# Filename : test.py
# author by : www.runoob.com
 
# 通过用户输入数字计算阶乘
 
# 获取用户输入的数字
num = 6789
factorial = 1
 
# 查看数字是负数，0 或 正数
if num < 0:
   print("抱歉，负数没有阶乘")
elif num == 0:
   print("0 的阶乘为 1")
else:
   for i in range(1,num + 1):
       factorial = factorial*i
   # print("%d 的阶乘为 %d" %(num,factorial))

string = "00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
print(len(string))