Dados dos números a y b . La tarea es restar b de a usando el método Complemento a 2 .
Nota : Números negativos representados como complemento a 2 de números positivos.
Por ejemplo, -5 se puede representar en forma binaria como Complemento a 2 de 5. Mira la imagen a continuación:
Ejemplos :
Input : a = 2, b = 3 Output : -1 Input : a = 9, b = 7 Output : 2
Para restar b de a . Escriba la expresión (ab) como:
(a - b) = a + (-b)
Ahora (-b) se puede escribir como (complemento a 2 de b). Entonces, la expresión anterior ahora se puede escribir como:
(a - b) = a + (2's complement of b)
Entonces, el problema ahora se reduce a «Sumar a al complemento a 2 de b «. La siguiente imagen ilustra el método de resta anterior para el primer ejemplo donde a = 2 y b = 3.
A continuación se muestra la implementación del método anterior:
C++
#include <bits/stdc++.h>
using namespace std;
// function to subtract two values
// using 2's complement method
int Subtract(int a, int b)
{
int c;
// ~b is the 1's Complement of b
// adding 1 to it make it 2's Complement
c = a + (~b + 1);
return c;
}
// Driver code
int main()
{
int a = 2, b = 3;
cout << Subtract(a, b)<<endl;
a = 9; b = 7;
cout << Subtract(a, b);
return 0;
}
Java
class GFG
{
// function to subtract two values
// using 2's complement method
static int Subtract(int a, int b)
{
int c;
// ~b is the 1's Complement
// of b adding 1 to it make
// it 2's Complement
c = a + (~b + 1);
return c;
}
// Driver code
public static void main(String[] args)
{
int a = 2, b = 3;
System.out.println(Subtract(a, b));
a = 9; b = 7;
System.out.println(Subtract(a, b));
}
}
// This code is contributed
// by ChitraNayal
Python3
# python3 program of subtraction of # two numbers using 2's complement . # function to subtract two values # using 2's complement method def Subtract(a,b): # ~b is the 1's Complement of b # adding 1 to it make it 2's Complement c = a + (~b + 1) return c # Driver code if __name__ == "__main__" : # multiple assignments a,b = 2,3 print(Subtract(a,b)) a,b = 9,7 print(Subtract(a,b))
C#
// C# program of subtraction of
// two numbers using 2's complement
using System;
class GFG
{
// function to subtract two values
// using 2's complement method
static int Subtract(int a, int b)
{
int c;
// ~b is the 1's Complement
// of b adding 1 to it make
// it 2's Complement
c = a + (~b + 1);
return c;
}
// Driver code
static void Main()
{
int a = 2, b = 3;
Console.WriteLine(Subtract(a, b));
a = 9; b = 7;
Console.WriteLine(Subtract(a, b));
}
}
// This code is contributed
// by mits
PHP
<?php
// function to subtract two values
// using 2's complement method
function Subtract($a, $b)
{
// ~b is the 1's Complement
// of b adding 1 to it make
// it 2's Complement
$c = $a + (~$b + 1);
return $c;
}
// Driver code
$a = 2;
$b = 3;
echo Subtract($a, $b) . "\n";
$a = 9;
$b = 7;
echo Subtract($a, $b) . "\n";
// This code is contributed
// by ChitraNayal
?>
Javascript
<script>
// Function to subtract two values
// using 2's complement method
function Subtract(a, b)
{
var c;
// ~b is the 1's Complement of b
// adding 1 to it make it 2's Complement
c = a + (~b + 1);
return c;
}
// Driver code
var a = 2, b = 3;
document.write( Subtract(a, b) + "<br>");
a = 9; b = 7;
document.write( Subtract(a, b));
// This code is contributed by itsok
</script>
Producción
-1 2
Complejidad del tiempo – O(nlog2(n))
Espacio auxiliar – O(1)
Método 2: enfoque básico o enfoque de fuerza bruta
Resta de dos números binarios, resta dos números binarios usando el método de complemento a 2.
Paso 1: Encuentra el complemento a 2 del sustraendo.
Paso 2: Sume el primer número y el complemento a 2 del sustraendo.
Paso 3: si se produce el acarreo, deséchelo. Si no hay acarreo, tome el complemento a 2 del resultado.
A continuación se muestra la implementación del enfoque anterior.
C++
//C++ code for above approach
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
// Program to subtract
void Subtract(int n, int a[],
int b[])
{
// 1's Complement
for(int i = 0; i < n; i++)
{
//Replace 1 by 0
if(b[i] == 1)
{
b[i] = 0;
}
//Replace 0 by 1
else
{
b[i] = 1;
}
}
//Add 1 at end to get 2's Complement
for(int i = n - 1; i >= 0; i--)
{
if(b[i] == 0)
{
b[i] = 1;
break;
}
else
{
b[i] = 0;
}
}
// Represents carry
int t = 0;
for(int i = n - 1; i >= 0; i--)
{
// Add a, b and carry
a[i] = a[i] + b[i] + t;
// If a[i] is 2
if(a[i] == 2)
{
a[i] = 0;
t = 1;
}
// If a[i] is 3
else if(a[i] == 3)
{
a[i] = 1;
t = 1;
}
else
t = 0;
}
cout << endl;
// If carry is generated
// discard the carry
if(t==1)
{
// print the result
for(int i = 0; i < n; i++)
{
// Print the result
cout<<a[i];
}
}
// if carry is not generated
else
{
// Calculate 2's Complement
// of the obtained result
for(int i = 0; i < n; i++)
{
if(a[i] == 1)
a[i] = 0;
else
a[i] = 1;
}
for(int i = n - 1; i >= 0; i--)
{
if(a[i] == 0)
{
a[i] = 1;
break;
}
else
a[i] = 0;
}
// Add -ve sign to represent
cout << "-";
// -ve result
// Print the resultant array
for(int i = 0; i < n; i++)
{
cout << a[i];
}
}
}
// Driver Code
int main()
{
int n;
n = 5;
int a[] = {1, 0, 1, 0, 1},
b[] = {1, 1, 0, 1, 0};
Subtract(n,a,b);
return 0;
}
Java
// Java code for above approach
import java.io.*;
class GFG{
// Program to subtract
static void Subtract(int n, int a[], int b[])
{
// 1's Complement
for(int i = 0; i < n; i++)
{
// Replace 1 by 0
if (b[i] == 1)
{
b[i] = 0;
}
// Replace 0 by 1
else
{
b[i] = 1;
}
}
// Add 1 at end to get 2's Complement
for(int i = n - 1; i >= 0; i--)
{
if (b[i] == 0)
{
b[i] = 1;
break;
}
else
{
b[i] = 0;
}
}
// Represents carry
int t = 0;
for(int i = n - 1; i >= 0; i--)
{
// Add a, b and carry
a[i] = a[i] + b[i] + t;
// If a[i] is 2
if (a[i] == 2)
{
a[i] = 0;
t = 1;
}
// If a[i] is 3
else if (a[i] == 3)
{
a[i] = 1;
t = 1;
}
else
t = 0;
}
System.out.println();
// If carry is generated
// discard the carry
if (t == 1)
{
// Print the result
for(int i = 0; i < n; i++)
{
// Print the result
System.out.print(a[i]);
}
}
// If carry is not generated
else
{
// Calculate 2's Complement
// of the obtained result
for(int i = 0; i < n; i++)
{
if (a[i] == 1)
a[i] = 0;
else
a[i] = 1;
}
for(int i = n - 1; i >= 0; i--)
{
if (a[i] == 0)
{
a[i] = 1;
break;
}
else
a[i] = 0;
}
// Add -ve sign to represent
System.out.print("-");
// -ve result
// Print the resultant array
for(int i = 0; i < n; i++)
{
System.out.print(a[i]);
}
}
}
// Driver Code
public static void main (String[] args)
{
int n;
n = 5;
int a[] = {1, 0, 1, 0, 1};
int b[] = {1, 1, 0, 1, 0};
Subtract(n, a, b);
}
}
// This code is contributed by avanitrachhadiya2155
Python3
# Python implementation of above approach
# Program to subtract
def Subtract(n, a, b):
# 1's Complement
for i in range(n):
# Replace 1 by 0
if (b[i] == 1):
b[i] = 0
# Replace 0 by 1
else:
b[i] = 1
# Add 1 at end to get 2's Complement
for i in range(n - 1, -1, -1):
if (b[i] == 0):
b[i] = 1
break
else:
b[i] = 0
# Represents carry
t = 0
for i in range(n - 1, -1, -1):
# Add a, b and carry
a[i] = a[i] + b[i] + t
# If a[i] is 2
if (a[i] == 2):
a[i] = 0
t = 1
# If a[i] is 3
elif (a[i] == 3):
a[i] = 1
t = 1
else:
t = 0
print()
# If carry is generated
# discard the carry
if (t == 1):
# Print the result
for i in range(n):
# Print the result
print(a[i],end="")
# If carry is not generated
else:
# Calculate 2's Complement
# of the obtained result
for i in range(n):
if (a[i] == 1):
a[i] = 0
else:
a[i] = 1
for i in range(n - 1, -1, -1):
if (a[i] == 0):
a[i] = 1
break
else:
a[i] = 0
# Add -ve sign to represent
print("-",end="")
# -ve result
# Print the resultant array
for i in range(n):
print(a[i],end="")
# Driver Code
n = 5
a=[1, 0, 1, 0, 1]
b=[1, 1, 0, 1, 0]
Subtract(n, a, b)
# This code is contributed by shinjanpatra
C#
// C# code for above approach
using System;
class GFG{
// Program to subtract
static void Subtract(int n, int[] a, int[] b)
{
// 1's Complement
for(int i = 0; i < n; i++)
{
// Replace 1 by 0
if (b[i] == 1)
{
b[i] = 0;
}
// Replace 0 by 1
else
{
b[i] = 1;
}
}
// Add 1 at end to get 2's Complement
for(int i = n - 1; i >= 0; i--)
{
if (b[i] == 0)
{
b[i] = 1;
break;
}
else
{
b[i] = 0;
}
}
// Represents carry
int t = 0;
for(int i = n - 1; i >= 0; i--)
{
// Add a, b and carry
a[i] = a[i] + b[i] + t;
// If a[i] is 2
if (a[i] == 2)
{
a[i] = 0;
t = 1;
}
// If a[i] is 3
else if (a[i] == 3)
{
a[i] = 1;
t = 1;
}
else
t = 0;
}
Console.WriteLine();
// If carry is generated
// discard the carry
if (t == 1)
{
// Print the result
for(int i = 0; i < n; i++)
{
// Print the result
Console.Write(a[i]);
}
}
// If carry is not generated
else
{
// Calculate 2's Complement
// of the obtained result
for(int i = 0; i < n; i++)
{
if (a[i] == 1)
a[i] = 0;
else
a[i] = 1;
}
for(int i = n - 1; i >= 0; i--)
{
if (a[i] == 0)
{
a[i] = 1;
break;
}
else
a[i] = 0;
}
// Add -ve sign to represent
Console.Write("-");
// -ve result
// Print the resultant array
for(int i = 0; i < n; i++)
{
Console.Write(a[i]);
}
}
}
// Driver Code
static public void Main()
{
int n;
n = 5;
int[] a = {1, 0, 1, 0, 1};
int[] b = {1, 1, 0, 1, 0};
Subtract(n, a, b);
}
}
// This code is contributed by rag2127
Javascript
<script>
// Javascript code for above approach
// Program to subtract
function Subtract(n,a,b)
{
// 1's Complement
for(let i = 0; i < n; i++)
{
// Replace 1 by 0
if (b[i] == 1)
{
b[i] = 0;
}
// Replace 0 by 1
else
{
b[i] = 1;
}
}
// Add 1 at end to get 2's Complement
for(let i = n - 1; i >= 0; i--)
{
if (b[i] == 0)
{
b[i] = 1;
break;
}
else
{
b[i] = 0;
}
}
// Represents carry
let t = 0;
for(let i = n - 1; i >= 0; i--)
{
// Add a, b and carry
a[i] = a[i] + b[i] + t;
// If a[i] is 2
if (a[i] == 2)
{
a[i] = 0;
t = 1;
}
// If a[i] is 3
else if (a[i] == 3)
{
a[i] = 1;
t = 1;
}
else
t = 0;
}
document.write("<br>");
// If carry is generated
// discard the carry
if (t == 1)
{
// Print the result
for(let i = 0; i < n; i++)
{
// Print the result
document.write(a[i]);
}
}
// If carry is not generated
else
{
// Calculate 2's Complement
// of the obtained result
for(let i = 0; i < n; i++)
{
if (a[i] == 1)
a[i] = 0;
else
a[i] = 1;
}
for(let i = n - 1; i >= 0; i--)
{
if (a[i] == 0)
{
a[i] = 1;
break;
}
else
a[i] = 0;
}
// Add -ve sign to represent
document.write("-");
// -ve result
// Print the resultant array
for(let i = 0; i < n; i++)
{
document.write(a[i]);
}
}
}
// Driver Code
let n = 5;
let a=[1, 0, 1, 0, 1];
let b=[1, 1, 0, 1, 0];
Subtract(n, a, b);
// This code is contributed by patel2127
</script>
Producción
-00101
Complejidad del tiempo : O(N)
Espacio Auxiliar : O(1)
Publicación traducida automáticamente
Artículo escrito por Naman_Garg y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA