给定以十进制为底的数字N，请以任意底数(底为b)查找其位数 _十进制

本文概述

C ++
Java
Python3
C#
的PHP

【给定以十进制为底的数字N，请以任意底数(底为b)查找其位数】给定以10为底的数字n, 请找到以b为底的数字位数。
限制条件：

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整个
例子：

Input : Number = 48 Base = 4 Output: 3 Explanation : (48)10 = (300)4Input : Number = 1446 Base = 7 Output: 4 Explanation : (446)10 = (4134)7

一种简单的方法: 将十进制数转换成给定的基数r，然后数位数。
一种有效的方法: 它驻留在该数的底数和该数的位数之间的关系。
通常：令n为正整数。底数

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代表

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具有

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如果数字

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, 如果是这样的话

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or

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因此, 以n为底的b表示形式的位数是

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在上面的方程中, 使用了基本变化对数性质。因此, 我们计算要计算位数的那个底数的对数。取其底值, 然后加1。
该想法可以进一步用于在基数r中找到基数b的给定数n的位数。所有要做的就是转换以10为底的数字, 然后应用上面的查找数字公式。当数字以10为底时, 计算任何底数的对数会更容易。
C ++

// C++ program to Find Number of digits // in base b. #include < iostream> #include < math.h> using namespace std; // function to print number of // digits void findNumberOfDigits( long n, int base) { // Calculating log using base // changing property and then // taking it floor and then // adding 1. int dig = ( int )( floor ( log (n) / log (base)) + 1); // printing output cout < < "The Number of digits of " < < "Number " < < n < < " in base " < < base < < " is " < < dig; }// Driver method int main() { // taking inputs long n = 1446; int base = 7; // calling the method findNumberOfDigits(n, base); return 0; }// This code is contributed by Manish Shaw // (manishshaw1)

Java

// Java program to Find Number // of digits in base b. class GFG {// function to print number of digits static void findNumberOfDigits( long n, int base) {// Calculating log using base changing // property and then taking it // floor and then adding 1. int dig = ( int )(Math.floor( Math.log(n) / Math.log(base)) + 1 ); // printing output System.out.println( "The Number of digits of Number " + n + " in base " + base + " is " + dig); }// Driver method public static void main(String[] args) { // taking inputs long n = 1446 ; int base = 7 ; // calling the method findNumberOfDigits(n, base); } }

Python3

# Python3 program to Find Number of digits # in base b.import math# function to print number of # digits def findNumberOfDigits(n, base):# Calculating log using base # changing property and then # taking it floor and then # adding 1. dig = (math.floor(math.log(n) / math.log(base)) + 1 )# printing output print ( "The Number of digits of" " Number {} in base {} is {}" . format (n, base, dig))# Driver method # taking inputs n = 1446 base = 7# calling the method findNumberOfDigits(n, base)# This code is contributed by # Manish Shaw (manishshaw1)

// C# program to Find Number of digits // in base b. using System; class GFG {// function to print number of // digits static void findNumberOfDigits( long n, int b) { // Calculating log using base // changing property and then // taking it floor and then // adding 1. int dig = ( int )(Math.Floor( Math.Log(n) / Math.Log(b)) + 1); // printing output Console.Write( "The Number of digits" + " of Number " + n + " in base " + b + " is " + dig); }// Driver method public static void Main() { // taking inputs long n = 1446; int b = 7; // calling the method findNumberOfDigits(n, b); } }// This code is contributed by Manish Shaw // (manishshaw1)

的PHP

< ?php // PHP program to Find Number // of digits in base b.// function to print // number of digits function findNumberOfDigits( $n , $b ) { // Calculating log using base // changing property and then // taking it floor and then // adding 1. $dig = (int)( floor (log( $n ) / log( $b )) + 1); // printing output echo ( "The Number of digits" . " of Number " . $n . " in base " . $b . " is " . $dig ); }// Driver Code $n = 1446; $b = 7; // calling the method findNumberOfDigits( $n , $b ); // This code is contributed by // Manish Shaw (manishshaw1) ?>

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