zoukankan      html  css  js  c++  java
  • HDU_1018_n(1e7)的阶乘的结果的位数

    http://acm.hdu.edu.cn/showproblem.php?pid=1018

    Big Number

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 33695    Accepted Submission(s): 15894


    Problem Description
    In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of digits in the factorial of the number.
     
    Input
    Input consists of several lines of integer numbers. The first line contains an integer n, which is the number of cases to be tested, followed by n lines, one integer 1 ≤ n ≤ 107 on each line.
     
    Output
    The output contains the number of digits in the factorial of the integers appearing in the input.
     
    Sample Input
    2 10 20
     
    Sample Output
    7 19
     
    任意一个正整数a的位数等于(int)log10(a) + 1;
    对于任意一个给定的正整数a,
      假设10^(x-1)<=a<10^x,那么显然a的位数为x位,
      又因为
      log10(10^(x-1))<=log10(a)<(log10(10^x))
      即x-1<=log10(a)<x
      则(int)log10(a)=x-1,
      即(int)log10(a)+1=x
      即a的位数是(int)log10(a)+1
    那么我们要求的就是
    (int)log10(A)+1,而:
    	log10(A)
            =log10(1*2*3*......n)  (根据log10(a*b) = log10(a) + log10(b)有)
            =log10(1)+log10(2)+log10(3)+......+log10(n)

    总结一下:n的阶乘的位数等于
    		  (int)(log10(1)+log10(2)+log10(3)+......+log10(n)) + 1
  • 相关阅读:
    ✨Synchronized底层实现---偏向锁
    🌞LCP 13. 寻宝
    ✨Synchronized底层实现---概述
    ⛅104. 二叉树的最大深度
    c++多线程之顺序调用类成员函数
    C++ STL实现总结
    C#小知识
    C#中HashTable和Dictionary的区别
    WPF的静态资源(StaticResource)和动态资源(DynamicResource)
    WPF之再谈MVVM
  • 原文地址:https://www.cnblogs.com/jasonlixuetao/p/5448892.html
Copyright © 2011-2022 走看看