Restar 1 sin operadores aritméticos

Escriba un programa para restar uno de un número dado. No se permite el uso de operadores como ‘+’, ‘-‘, ‘*’, ‘/’, ‘++’, ‘–’…etc. 

Ejemplos: 

Input:  12
Output: 11

Input:  6
Output: 5

Método 1 
Para restar 1 de un número x (por ejemplo, 0011001000), voltea todos los bits después del 1 bit más a la derecha (obtenemos 001100 1 111). Finalmente, cambie también el bit más a la derecha (obtenemos 0011000111) para obtener la respuesta.

// C++ code to subtract
// one from a given number
#include <iostream>
using namespace std;
 
int subtractOne(int x)
{
    int m = 1;
 
    // Flip all the set bits
    // until we find a 1
    while (!(x & m))
    {
        x = x ^ m;
        m <<= 1;
    }
 
    // Flip the rightmost 1 bit
    x = x ^ m;
    return x;
}
 
// Driver code
int main()
{
    cout << subtractOne(13) << endl;
    return 0;
}
 
// This code is contributed by noob2000

// C code to subtract
// one from a given number
#include <stdio.h>
 
int subtractOne(int x)
{
    int m = 1;
 
    // Flip all the set bits
    // until we find a 1
    while (!(x & m)) {
        x = x ^ m;
        m <<= 1;
    }
 
    // flip the rightmost 1 bit
    x = x ^ m;
    return x;
}
 
/* Driver program to test above functions*/
int main()
{
    printf("%d", subtractOne(13));
    return 0;
}

// Java code to subtract
// one from a given number
import java.io.*;
 
class GFG
{
static int subtractOne(int x)
{
    int m = 1;
 
    // Flip all the set bits
    // until we find a 1
    while (!((x & m) > 0))
    {
        x = x ^ m;
        m <<= 1;
    }
 
    // flip the rightmost
    // 1 bit
    x = x ^ m;
    return x;
}
 
// Driver Code
public static void main (String[] args)
{
    System.out.println(subtractOne(13));
}
}
 
// This code is contributed
// by anuj_67.

# Python 3 code to subtract one from
# a given number
def subtractOne(x):
    m = 1
 
    # Flip all the set bits
    # until we find a 1
    while ((x & m) == False):
        x = x ^ m
        m = m << 1
     
    # flip the rightmost 1 bit
    x = x ^ m
    return x
 
# Driver Code
if __name__ == '__main__':
    print(subtractOne(13))
     
# This code is contributed by
# Surendra_Gangwar

// C# code to subtract
// one from a given number
using System;
 
class GFG
{
static int subtractOne(int x)
{
    int m = 1;
 
    // Flip all the set bits
    // until we find a 1
    while (!((x & m) > 0))
    {
        x = x ^ m;
        m <<= 1;
    }
 
    // flip the rightmost
    // 1 bit
    x = x ^ m;
    return x;
}
 
// Driver Code
public static void Main ()
{
    Console.WriteLine(subtractOne(13));
}
}
 
// This code is contributed
// by anuj_67.

<?php
// PHP code to subtract
// one from a given number
 
function subtractOne($x)
{
    $m = 1;
 
    // Flip all the set bits
    // until we find a 1
    while (!($x & $m))
    {
        $x = $x ^ $m;
        $m <<= 1;
    }
 
    // flip the
    // rightmost 1 bit
    $x = $x ^ $m;
    return $x;
}
 
// Driver Code
    echo subtractOne(13);
     
// This code is contributed
// by anuj_67.
?>

<script>
 
// JavaScript code to subtract
// one from a given number
 
function subtractOne(x)
{
    let m = 1;
 
    // Flip all the set bits
    // until we find a 1
    while (!(x & m)) {
        x = x ^ m;
        m <<= 1;
    }
 
    // flip the rightmost 1 bit
    x = x ^ m;
    return x;
}
 
/* Driver program to test above functions*/
 
    document.write(subtractOne(13));
 
 
// This code is contributed by Surbhi Tyagi.
 
</script>

12

Método 2 (si se permite +) 
Sabemos que el número negativo se representa en forma de complemento a 2 en la mayoría de las arquitecturas. Tenemos el siguiente lema válido para la representación en complemento a 2 de números con signo.
Digamos que x es el valor numérico de un número, entonces
~x = -(x+1) [ ~ es para complemento bit a bit ]
Sumar 2x en ambos lados, 
2x + ~x = x – 1
Para obtener 2x, desplaza x a la izquierda una vez . 

#include <bits/stdc++.h>
using namespace std;
 
int subtractOne(int x) { return ((x << 1) + (~x)); }
 
/* Driver program to test above functions*/
int main()
{
    cout<< subtractOne(13);
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

#include <stdio.h>
 
int subtractOne(int x) { return ((x << 1) + (~x)); }
 
/* Driver program to test above functions*/
int main()
{
    printf("%d", subtractOne(13));
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

class GFG
{
 
    static int subtractOne(int x)
    {
        return ((x << 1) + (~x));
    }
 
    /* Driver code*/
    public static void main(String[] args)
    {
        System.out.printf("%d", subtractOne(13));
    }
}
 
// This code has been contributed by 29AjayKumar

def subtractOne(x):
 
    return ((x << 1) + (~x));
 
# Driver code
print(subtractOne(13));
 
# This code is contributed by mits

using System;
     
class GFG
{
 
    static int subtractOne(int x)
    {
        return ((x << 1) + (~x));
    }
 
    /* Driver code*/
    public static void Main(String[] args)
    {
        Console.Write("{0}", subtractOne(13));
    }
}
 
// This code contributed by Rajput-Ji

<?php
function subtractOne($x)
{
    return (($x << 1) + (~$x));
}
 
/* Driver code*/
 
print(subtractOne(13));
 
// This code has been contributed by mits
?>

<script>
 
function subtractOne(x)
{
    return ((x << 1) + (~x));
}
 
/* Driver program to test above functions*/
 
    document.write((subtractOne(13)));
 
// This is code is contributed by Mayank Tyagi
 
</script>

12