El problema es convertir el número binario dado (representado como string) a su número octal equivalente. La entrada podría ser muy grande y es posible que no encaje ni siquiera en int largo largo sin firmar.
Ejemplos:
Input : 110001110 Output : 616 Input : 1111001010010100001.010110110011011 Output : 1712241.26633
La idea es considerar la entrada binaria como una string de caracteres y luego seguir los pasos:
- Obtenga la longitud de la substring a la izquierda y a la derecha del punto decimal (‘.’) como left_len y right_len .
- Si left_len no es un múltiplo de 3, agregue un número mínimo de 0 al principio para que la longitud de la substring izquierda sea un múltiplo de 3.
- Si right_len no es un múltiplo de 3, agregue un número mínimo de 0 al final para que la longitud de la substring derecha sea un múltiplo de 3.
- Ahora, desde la izquierda, extraiga una por una las substrings de longitud 3 y agregue su código octal correspondiente al resultado.
- Si se encuentra un decimal (‘.’) en el medio, agréguelo al resultado.
C++
// C++ implementation to convert a binary number
// to octal number
#include <bits/stdc++.h>
using namespace std;
// function to create map between binary
// number and its equivalent octal
void createMap(unordered_map<string, char> *um)
{
(*um)["000"] = '0';
(*um)["001"] = '1';
(*um)["010"] = '2';
(*um)["011"] = '3';
(*um)["100"] = '4';
(*um)["101"] = '5';
(*um)["110"] = '6';
(*um)["111"] = '7';
}
// Function to find octal equivalent of binary
string convertBinToOct(string bin)
{
int l = bin.size();
int t = bin.find_first_of('.');
// length of string before '.'
int len_left = t != -1 ? t : l;
// add min 0's in the beginning to make
// left substring length divisible by 3
for (int i = 1; i <= (3 - len_left % 3) % 3; i++)
bin = '0' + bin;
// if decimal point exists
if (t != -1)
{
// length of string after '.'
int len_right = l - len_left - 1;
// add min 0's in the end to make right
// substring length divisible by 3
for (int i = 1; i <= (3 - len_right % 3) % 3; i++)
bin = bin + '0';
}
// create map between binary and its
// equivalent octal code
unordered_map<string, char> bin_oct_map;
createMap(&bin_oct_map);
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;
// if '.' is encountered add it to result
if (bin.at(i) == '.')
{
octal += '.';
i++;
}
}
// required octal number
return octal;
}
// Driver program to test above
int main()
{
string bin = "1111001010010100001.010110110011011";
cout << "Octal number = "
<< convertBinToOct(bin);
return 0;
}
Java
// Java implementation to convert a
// binary number to octal number
import java.io.*;
import java.util.*;
class GFG{
// Function to create map between binary
// number and its equivalent hexadecimal
static void createMap(Map<String, Character> um)
{
um.put("000", '0');
um.put("001", '1');
um.put("010", '2');
um.put("011", '3');
um.put("100", '4');
um.put("101", '5');
um.put("110", '6');
um.put("111", '7');
}
// Function to find octal equivalent of binary
static String convertBinToOct(String bin)
{
int l = bin.length();
int t = bin.indexOf('.');
// Length of string before '.'
int len_left = t != -1 ? t : l;
// Add min 0's in the beginning to make
// left substring length divisible by 3
for(int i = 1;
i <= (3 - len_left % 3) % 3;
i++)
bin = '0' + bin;
// If decimal point exists
if (t != -1)
{
// Length of string after '.'
int len_right = l - len_left - 1;
// add min 0's in the end to make right
// substring length divisible by 3
for(int i = 1;
i <= (3 - len_right % 3) % 3;
i++)
bin = bin + '0';
}
// Create map between binary and its
// equivalent octal code
Map<String,
Character> bin_oct_map = new HashMap<String,
Character>();
createMap(bin_oct_map);
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;
// If '.' is encountered add it to result
if (bin.charAt(i) == '.')
{
octal += '.';
i++;
}
}
// Required octal number
return octal;
}
// Driver code
public static void main(String[] args)
{
String bin = "1111001010010100001.010110110011011";
System.out.println("Octal number = " +
convertBinToOct(bin));
}
}
// This code is contributed by jithin
Python3
# Python3 implementation to convert a binary number
# to octal number
# function to create map between binary
# number and its equivalent octal
def createMap(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'
# Function to find octal equivalent of binary
def convertBinToOct(bin):
l = len(bin)
# length of string before '.'
t = -1
if '.' in bin:
t = bin.index('.')
len_left = t
else:
len_left = l
# add min 0's in the beginning to make
# left substring length divisible by 3
for i in range(1, (3 - len_left % 3) % 3 + 1):
bin = '0' + bin
# if decimal point exists
if (t != -1):
# length of string after '.'
len_right = l - len_left - 1
# add min 0's in the end to make right
# substring length divisible by 3
for i in range(1, (3 - len_right % 3) % 3 + 1):
bin = bin + '0'
# create dictionary between binary and its
# equivalent octal code
bin_oct_map = {}
createMap(bin_oct_map)
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
# if '.' is encountered add it to result
if (bin[i] == '.'):
octal += '.'
i += 1
# required octal number
return octal
# Driver Code
bin = "1111001010010100001.010110110011011"
print("Octal number = ",
convertBinToOct(bin))
# This code is contributed
# by Atul_kumar_Shrivastava
C#
// C# implementation to convert a
// binary number to octal number
using System;
using System.Collections.Generic;
public class GFG{
// Function to create map between binary
// number and its equivalent hexadecimal
static void createMap(Dictionary<String, char> um)
{
um.Add("000", '0');
um.Add("001", '1');
um.Add("010", '2');
um.Add("011", '3');
um.Add("100", '4');
um.Add("101", '5');
um.Add("110", '6');
um.Add("111", '7');
}
// Function to find octal equivalent of binary
static String convertBinToOct(String bin)
{
int l = bin.Length;
int t = bin.IndexOf('.');
int i = 0;
// Length of string before '.'
int len_left = t != -1 ? t : l;
// Add min 0's in the beginning to make
// left substring length divisible by 3
for(i = 1;
i <= (3 - len_left % 3) % 3;
i++)
bin = '0' + bin;
// If decimal point exists
if (t != -1)
{
// Length of string after '.'
int len_right = l - len_left - 1;
// add min 0's in the end to make right
// substring length divisible by 3
for(i = 1;
i <= (3 - len_right % 3) % 3;
i++)
bin = bin + '0';
}
// Create map between binary and its
// equivalent octal code
Dictionary<String,
char> bin_oct_map = new Dictionary<String,
char>();
createMap(bin_oct_map);
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;
// If '.' is encountered add it to result
if (bin[i] == '.')
{
octal += '.';
i++;
}
}
// Required octal number
return octal;
}
// Driver code
public static void Main(String[] args)
{
String bin = "1111001010010100001.010110110011011";
Console.WriteLine("Octal number = " +
convertBinToOct(bin));
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// Javascript implementation to convert a
// binary number to octal number
// Function to create map between binary
// number and its equivalent hexadecimal
function createMap(um)
{
um.set("000", '0');
um.set("001", '1');
um.set("010", '2');
um.set("011", '3');
um.set("100", '4');
um.set("101", '5');
um.set("110", '6');
um.set("111", '7');
}
// Function to find octal equivalent of binary
function convertBinToOct(bin)
{
let l = bin.length;
let t = bin.indexOf('.');
// Length of string before '.'
let len_left = t != -1 ? t : l;
// Add min 0's in the beginning to make
// left substring length divisible by 3
for(let i = 1;
i <= (3 - len_left % 3) % 3;
i++)
bin = '0' + bin;
// If decimal point exists
if (t != -1)
{
// Length of string after '.'
let len_right = l - len_left - 1;
// add min 0's in the end to make right
// substring length divisible by 3
for(let i = 1;
i <= (3 - len_right % 3) % 3;
i++)
bin = bin + '0';
}
// Create map between binary and its
// equivalent octal code
let bin_oct_map = new Map();
createMap(bin_oct_map);
let i = 0;
let octal = "";
while (true)
{
// One by one extract from left, substring
// of size 3 and add its octal code
octal += bin_oct_map.get(bin.substr(i, 3));
i += 3;
if (i == bin.length)
break;
// If '.' is encountered add it to result
if (bin.charAt(i) == '.')
{
octal += '.';
i++;
}
}
// Required octal number
return octal;
}
// Driver code
let bin = "1111001010010100001.010110110011011";
document.write("Octal number = " +
convertBinToOct(bin));
// This code is contributed by gfgking
</script>
Producción:
Octal number = 1712241.26633
Complejidad de tiempo: O(n), donde n es la longitud de la string.
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA