Conversión de decimal a octal con un uso mínimo de operadores aritméticos

Dado un número decimal n sin punto flotante. El problema es convertir el número decimal en número octal con un uso mínimo de operadores aritméticos. 

Ejemplos: 

Input : n = 10
Output : 12 
12 is octal equivalent of decimal 10.

Input : n = 151
Output : 227

Enfoque: Los siguientes son los pasos:

1. Realice la conversión de decimal a binario sin usar operadores aritméticos del número dado n . Consulte esta publicación. Sea este número bin .
2. Convierta el contenedor de números binarios a octal. Consulte esta publicación.

// C++ implementation of decimal to octal conversion
// with minimum use of arithmetic operators
#include <bits/stdc++.h>
 
using namespace std;
 
// function for decimal to binary conversion
// without using arithmetic operators
string decToBin(int n)
{
    if (n == 0)
        return "0";
      
    // to store the binary equivalent of decimal
    string bin = "";   
    while (n > 0)   
    {
        // to get the last binary digit of the
        // number 'n' and accumulate it at the
        // beginning of 'bin'
        bin = ((n & 1) == 0 ? '0' : '1') + bin;
          
        // right shift 'n' by 1
        n >>= 1;
    }
      
    // required binary number
    return bin;
}
 
  
// Function to find octal equivalent of binary
string convertBinToOct(string bin)
{
    int l = bin.size();
          
    // add min 0's in the beginning to make
    // string length divisible by 3
    for (int i = 1; i <= (3 - l % 3) % 3; i++)
        bin = '0' + bin;
       
    // create map between binary and its
    // equivalent octal code
    unordered_map<string, char> bin_oct_map;
    bin_oct_map["000"] = '0';
    bin_oct_map["001"] = '1';
    bin_oct_map["010"] = '2';
    bin_oct_map["011"] = '3';
    bin_oct_map["100"] = '4';
    bin_oct_map["101"] = '5';
    bin_oct_map["110"] = '6';
    bin_oct_map["111"] = '7'; 
       
    int i = 0;
    string octal = "";     
    while (1)
    {
        // one by one extract from left, substring
        // of size 3 and add its octal code
        octal += bin_oct_map[bin.substr(i, 3)];
        i += 3;
        if (i == bin.size())
            break;       
    }
       
    // required octal number
    return octal;   
}
 
// function to find octal equivalent of decimal
string decToOctal(int n)
{
    // convert decimal to binary
    string bin = decToBin(n);
     
    // convert binary to octal
    // required octal equivalent of decimal
    return convertBinToOct(bin);
}
 
// Driver program to test above
int main()
{
    int n = 151;
    cout << decToOctal(n);
    return 0;
}

// Java implementation of decimal to octal
// conversion with minimum use of
// arithmetic operators
import java.util.*;
 
class GFG
{
 
// function for decimal to binary conversion
// without using arithmetic operators
static String decToBin(int n)
{
    if (n == 0)
        return "0";
     
    // to store the binary equivalent of decimal
    String bin = "";
    while (n > 0)
    {
        // to get the last binary digit of the
        // number 'n' and accumulate it at the
        // beginning of 'bin'
        bin = ((n & 1) == 0 ? '0' : '1') + bin;
         
        // right shift 'n' by 1
        n >>= 1;
    }
     
    // required binary number
    return bin;
}
 
// Function to find octal equivalent of binary
static String convertBinToOct(String bin)
{
    int l = bin.length();
         
    // add min 0's in the beginning to make
    // string length divisible by 3
    for (int i = 1; i <= (3 - l % 3) % 3; i++)
        bin = '0' + bin;
         
    // create map between binary and its
    // equivalent octal code
    Map<String,
        Character> bin_oct_map = new HashMap<>();
    bin_oct_map.put("000", '0');
    bin_oct_map.put("001", '1');
    bin_oct_map.put("010", '2');
    bin_oct_map.put("011", '3');
    bin_oct_map.put("100", '4');
    bin_oct_map.put("101", '5');
    bin_oct_map.put("110", '6');
    bin_oct_map.put("111", '7');
         
    int i = 0;
    String octal = "";    
    while (true)
    {
        // one by one extract from left, substring
        // of size 3 and add its octal code
        octal += bin_oct_map.get(bin.substring(i, i + 3));
        i += 3;
        if (i == bin.length())
            break;    
    }
         
    // required octal number
    return octal;
}
 
// function to find octal equivalent of decimal
static String decToOctal(int n)
{
    // convert decimal to binary
    String bin = decToBin(n);
     
    // convert binary to octal
    // required octal equivalent of decimal
    return convertBinToOct(bin);
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 151;
    System.out.println(decToOctal(n));
}
}
 
// This code is contributed by Rajput-Ji

# Python3 implementation of decimal to octal conversion
# with minimum use of arithmetic operators
 
# function for decimal to binary conversion
# without using arithmetic operators
def decToBin(n):
 
    if (n == 0):
        return "0"
 
    # to store the binary equivalent of decimal
    bin = ""
     
    while (n > 0):
     
        # to get the last binary digit of the
        # number 'n' and accumulate it at the
        # beginning of 'bin'
        bin = ('0' if(n & 1) == 0 else '1') + bin
 
        # right shift 'n' by 1
        n >>= 1
 
    # required binary number
    return bin
 
# Function to find octal equivalent of binary
def convertBinToOct(bin):
 
    l = len(bin)
 
    # add min 0's in the beginning to make
    # string length divisible by 3
    for i in range(1,((3 - l % 3) % 3) + 1):
        bin = '0' + bin
 
    # create map between binary and its
    # equivalent octal code
     
    bin_oct_map = dict()
     
    bin_oct_map["000"] = '0'
    bin_oct_map["001"] = '1'
    bin_oct_map["010"] = '2'
    bin_oct_map["011"] = '3'
    bin_oct_map["100"] = '4'
    bin_oct_map["101"] = '5'
    bin_oct_map["110"] = '6'
    bin_oct_map["111"] = '7'
 
    i = 0
    octal = ""
     
    while (True):
     
        # one by one extract from left, substring
        # of size 3 and add its octal code
        octal += bin_oct_map[bin[i:i + 3]]
        i += 3
        if (i == len(bin)):
            break
     
    # required octal number
    return octal
 
# function to find octal equivalent of decimal
def decToOctal(n):
 
    # convert decimal to binary
    bin = decToBin(n)
 
    # convert binary to octal
    # required octal equivalent of decimal
    return convertBinToOct(bin)
 
# Driver program to test above
if __name__=='__main__':
 
    n = 151
    print(decToOctal(n))
     
# This code is contributed by pratham76

// C# implementation of decimal to octal
// conversion with minimum use of
// arithmetic operators
using System;
using System.Collections.Generic;
 
class GFG{
  
// Function for decimal to binary
// conversion without using
// arithmetic operators
static string decToBin(int n)
{
    if (n == 0)
        return "0";
      
    // To store the binary equivalent
    // of decimal
    string bin = "";
    while (n > 0)
    {
         
        // To get the last binary digit of the
        // number 'n' and accumulate it at the
        // beginning of 'bin'
        bin = ((n & 1) == 0 ? '0' : '1') + bin;
          
        // Right shift 'n' by 1
        n >>= 1;
    }
     
    // Required binary number
    return bin;
}
  
// Function to find octal equivalent of binary
static string convertBinToOct(string bin)
{
    int l = bin.Length;
          
    // Add min 0's in the beginning to make
    // string length divisible by 3
    for(int j = 1; j <= (3 - l % 3) % 3; j++)
        bin = '0' + bin;
          
    // Create map between binary and its
    // equivalent octal code
    Dictionary<string,
               char> bin_oct_map = new Dictionary<string,
                                                  char>();
     
    bin_oct_map.Add("000", '0');
    bin_oct_map.Add("001", '1');
    bin_oct_map.Add("010", '2');
    bin_oct_map.Add("011", '3');
    bin_oct_map.Add("100", '4');
    bin_oct_map.Add("101", '5');
    bin_oct_map.Add("110", '6');
    bin_oct_map.Add("111", '7');
          
    int i = 0;
    string octal = ""; 
     
    while (true)
    {
         
        // One by one extract from left, substring
        // of size 3 and add its octal code
        octal += bin_oct_map[bin.Substring(i, 3)];
        i += 3;
         
        if (i == bin.Length)
            break;    
    }
          
    // Required octal number
    return octal;
}
  
// Function to find octal equivalent of decimal
static string decToOctal(int n)
{
     
    // Convert decimal to binary
    string bin = decToBin(n);
      
    // Convert binary to octal
    // required octal equivalent
    // of decimal
    return convertBinToOct(bin);
}
  
// Driver Code
public static void Main(string[] args)
{
    int n = 151;
     
    Console.Write(decToOctal(n));
}
}
 
// This code is contributed by rutvik_56

<script>
 
// Javascript implementation of decimal to octal conversion
// with minimum use of arithmetic operators
 
// function for decimal to binary conversion
// without using arithmetic operators
function decToBin(n)
{
    if (n == 0)
        return "0";
      
    // to store the binary equivalent of decimal
    var bin = "";   
    while (n > 0)   
    {
        // to get the last binary digit of the
        // number 'n' and accumulate it at the
        // beginning of 'bin'
        bin = ((n & 1) == 0 ? '0' : '1') + bin;
          
        // right shift 'n' by 1
        n >>= 1;
    }
      
    // required binary number
    return bin;
}
 
  
// Function to find octal equivalent of binary
function convertBinToOct(bin)
{
    var l = bin.length;
          
    // add min 0's in the beginning to make
    // string length divisible by 3
    for (var i = 1; i <= (3 - l % 3) % 3; i++)
        bin = '0' + bin;
       
    // create map between binary and its
    // equivalent octal code
    var bin_oct_map = new Map();
    bin_oct_map.set("000",'0');
    bin_oct_map.set("001",'1');
    bin_oct_map.set("010",'2');
    bin_oct_map.set("011",'3');
    bin_oct_map.set("100",'4');
    bin_oct_map.set("101",'5');
    bin_oct_map.set("110",'6');
    bin_oct_map.set("111",'7');
       
    var i = 0;
    var octal = "";     
    while (1)
    {
        // one by one extract from left, substring
        // of size 3 and add its octal code
        octal += bin_oct_map.get(bin.substring(i, i+3));
        i += 3;
        if (i == bin.length)
            break;       
    }
       
    // required octal number
    return octal;   
}
 
// function to find octal equivalent of decimal
function decToOctal(n)
{
    // convert decimal to binary
    var bin = decToBin(n);
     
    // convert binary to octal
    // required octal equivalent of decimal
    return convertBinToOct(bin);
}
 
// Driver program to test above
var n = 151;
document.write( decToOctal(n));ing
 
// This code is contributed by itsok.
</script>

Producción: 

227

Complejidad de tiempo: O(n), donde n es la longitud de la string binaria.