实验吧 手脑并用
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))
知识要大家一起分享,但带上原文链接是对作者的尊重。