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  • 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))
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  • 原文地址:https://www.cnblogs.com/zhaijiahui/p/7705371.html
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