编写一个函数减法（x，y），返回x-y，其中x和y是整数。该函数不应使用任何算术运算符（+、++、–、…等）。 这个想法是使用位运算符。 使用位运算符讨论了两个数的加法 .就像加法一样，这个想法是使用 减法器 思维方式
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下面给出了半减法器的真值表。
X Y Diff Borrow0 0 0 00 1 1 11 0 1 01 1 0 0
从上表中，我们可以绘制“差异”和“借用”的卡诺图。 所以，逻辑方程是：
Diff = y ⊕ x Borrow = x' . y
资料来源： Wikipedia减法器页面
以下是基于上述等式的实现。
C++
// C++ program to Subtract two numbers // without using arithmetic operators #include <iostream> using namespace std; int subtract( int x, int y) { // Iterate till there // is no carry while (y != 0) { // borrow contains common // set bits of y and unset // bits of x int borrow = (~x) & y; // Subtraction of bits of x // and y where at least one // of the bits is not set x = x ^ y; // Borrow is shifted by one // so that subtracting it from // x gives the required sum y = borrow << 1; } return x; } // Driver Code int main() { int x = 29, y = 13; cout << "x - y is " << subtract(x, y); return 0; } // This code is contributed by shivanisinghss2110 |
C
// C program to Subtract two numbers // without using arithmetic operators #include<stdio.h> int subtract( int x, int y) { // Iterate till there // is no carry while (y != 0) { // borrow contains common // set bits of y and unset // bits of x int borrow = (~x) & y; // Subtraction of bits of x // and y where at least one // of the bits is not set x = x ^ y; // Borrow is shifted by one // so that subtracting it from // x gives the required sum y = borrow << 1; } return x; } // Driver Code int main() { int x = 29, y = 13; printf ( "x - y is %d" , subtract(x, y)); return 0; } |
JAVA
// Java Program to subtract two Number // without using arithmetic operator import java.io.*; class GFG { static int subtract( int x, int y) { // Iterate till there // is no carry while (y != 0 ) { // borrow contains common // set bits of y and unset // bits of x int borrow = (~x) & y; // Subtraction of bits of x // and y where at least one // of the bits is not set x = x ^ y; // Borrow is shifted by one // so that subtracting it from // x gives the required sum y = borrow << 1 ; } return x; } // Driver Code public static void main (String[] args) { int x = 29 , y = 13 ; System.out.println( "x - y is " + subtract(x, y)); } } // This code is contributed by vt_m |
Python3
def subtract(x, y): # Iterate till there # is no carry while (y ! = 0 ): # borrow contains common # set bits of y and unset # bits of x borrow = (~x) & y # Subtraction of bits of x # and y where at least one # of the bits is not set x = x ^ y # Borrow is shifted by one # so that subtracting it from # x gives the required sum y = borrow << 1 return x # Driver Code x = 29 y = 13 print ( "x - y is" ,subtract(x, y)) # This code is contributed by # Smitha Dinesh Semwal |
C#
// C# Program to subtract two Number // without using arithmetic operator using System; class GFG { static int subtract( int x, int y) { // Iterate till there // is no carry while (y != 0) { // borrow contains common // set bits of y and unset // bits of x int borrow = (~x) & y; // Subtraction of bits of x // and y where at least one // of the bits is not set x = x ^ y; // Borrow is shifted by one // so that subtracting it from // x gives the required sum y = borrow << 1; } return x; } // Driver Code public static void Main () { int x = 29, y = 13; Console.WriteLine( "x - y is " + subtract(x, y)); } } // This code is contributed by anuj_67. |
PHP
<?php // PHP Program to subtract two Number // without using arithmetic operator function subtract( $x , $y ) { // Iterate till there is no carry while ( $y != 0) { // borrow contains common set // bits of y and unset // bits of x $borrow = (~ $x ) & $y ; // Subtraction of bits of x // and y where at least // one of the bits is not set $x = $x ^ $y ; // Borrow is shifted by one so // that subtracting it from // x gives the required sum $y = $borrow << 1; } return $x ; } // Driver Code $x = 29; $y = 13; echo "x - y is " , subtract( $x , $y ); // This code is contributed by Ajit ?> |
Javascript
<script> // JavaScript program to Subtract two numbers // without using arithmetic operators function subtract(x, y) { // Iterate till there // is no carry while (y != 0) { // borrow contains common // set bits of y and unset // bits of x let borrow = (~x) & y; // Subtraction of bits of x // and y where at least one // of the bits is not set x = x ^ y; // Borrow is shifted by one // so that subtracting it from // x gives the required sum y = borrow << 1; } return x; } // Driver Code let x = 29, y = 13; document.write( "x - y is " + subtract(x, y)); // This code is contributed by Surbhi Tyagi. </script> |
输出：
x - y is 16
时间复杂度：O（logy）
辅助空间：O（1）
下面是相同方法的递归实现。
C++
#include <iostream> using namespace std; int subtract( int x, int y) { if (y == 0) return x; return subtract(x ^ y, (~x & y) << 1); } // Driver program int main() { int x = 29, y = 13; cout << "x - y is " << subtract(x, y); return 0; } // this code is contributed by shivanisinghss2110 |
C
#include<stdio.h> int subtract( int x, int y) { if (y == 0) return x; return subtract(x ^ y, (~x & y) << 1); } // Driver program int main() { int x = 29, y = 13; printf ( "x - y is %d" , subtract(x, y)); return 0; } |
JAVA
// Java Program to subtract two Number // without using arithmetic operator // Recursive implementation. class GFG { static int subtract( int x, int y) { if (y == 0 ) return x; return subtract(x ^ y, (~x & y) << 1 ); } // Driver program public static void main(String[] args) { int x = 29 , y = 13 ; System.out.printf( "x - y is %d" , subtract(x, y)); } } // This code is contributed by // Smitha Dinesh Semwal. |
Python3
# Python Program to # subtract two Number # without using arithmetic operator # Recursive implementation. def subtract(x, y): if (y = = 0 ): return x return subtract(x ^ y, (~x & y) << 1 ) # Driver program x = 29 y = 13 print ( "x - y is" , subtract(x, y)) # This code is contributed by # Smitha Dinesh Semwal |
C#
// C# Program to subtract two Number // without using arithmetic operator // Recursive implementation. using System; class GFG { static int subtract( int x, int y) { if (y == 0) return x; return subtract(x ^ y, (~x & y) << 1); } // Driver program public static void Main() { int x = 29, y = 13; Console.WriteLine( "x - y is " + subtract(x, y)); } } // This code is contributed by anuj_67. |
PHP
<?php function subtract( $x , $y ) { if ( $y == 0) return $x ; return subtract( $x ^ $y , (~ $x & $y ) << 1); } // Driver Code $x = 29; $y = 13; echo "x - y is " , subtract( $x , $y ); # This code is contributed by ajit ?> |
Javascript
<script> // javascript Program to subtract two Number // without using arithmetic operator // Recursive implementation. function subtract(x , y) { if (y == 0) return x; return subtract(x ^ y, (~x & y) << 1); } // Driver program var x = 29, y = 13; document.write( "x - y is " + subtract(x, y)); // This code is contributed by Princi Singh </script> |
输出：
x - y is 16
时间复杂度：O（logy）
辅助空间：O（对数y）
这篇文章是有贡献的 Dheeraj 。如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请发表评论
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