算法设计（如何用移位运算符将两个数相乘（）） _移位运算将两个数相乘

本文概述

C++
Java
Python3
C#
PHP

【算法设计（如何用移位运算符将两个数相乘（））】对于给定的两个数字n和m, 你必须找到n * m而不使用任何乘法运算符。
例子：

Input: n = 25 , m = 13 Output: 325Input: n = 50 , m = 16 Output: 800

我们可以使用移位运算符解决此问题。这个想法基于这样一个事实, 即每个数字都可以二进制形式表示。与数字的乘积等效于乘以2的幂。可以使用左移位运算符获得2的幂。
检查m的二进制表示形式中的每个设置位, 以及每个左移n的设置位, 计数次数, 以计算m的设置位的位置值并将该值相加。
C++

// CPP program to find multiplication // of two number without use of // multiplication operator #include< bits/stdc++.h> using namespace std; // Function for multiplication int multiply( int n, int m) { int ans = 0, count = 0; while (m) { // check for set bit and left // shift n, count times if (m % 2 == 1) ans += n < < count; // increment of place value (count) count++; m /= 2; } return ans; }// Driver code int main() { int n = 20 , m = 13; cout < < multiply(n, m); return 0; }

Java

// Java program to find multiplication // of two number without use of // multiplication operator class GFG {// Function for multiplication static int multiply( int n, int m) { int ans = 0 , count = 0 ; while (m > 0 ) { // check for set bit and left // shift n, count times if (m % 2 == 1 ) ans += n < < count; // increment of place // value (count) count++; m /= 2 ; }return ans; }// Driver code public static void main (String[] args) { int n = 20 , m = 13 ; System.out.print( multiply(n, m) ); } }// This code is contributed by Anant Agarwal.

Python3

# python 3 program to find multiplication # of two number without use of # multiplication operator# Function for multiplication def multiply(n, m): ans = 0 count = 0 while (m): # check for set bit and left # shift n, count times if (m % 2 = = 1 ): ans + = n < < count# increment of place value (count) count + = 1 m = int (m / 2 )return ans# Driver code if __name__ = = '__main__' : n = 20 m = 13 print (multiply(n, m))# This code is contributed by # Ssanjit_Prasad

// C# program to find multiplication // of two number without use of // multiplication operator using System; class GFG {// Function for multiplication static int multiply( int n, int m) { int ans = 0, count = 0; while (m > 0) { // check for set bit and left // shift n, count times if (m % 2 == 1) ans += n < < count; // increment of place // value (count) count++; m /= 2; }return ans; }// Driver Code public static void Main () { int n = 20, m = 13; Console.WriteLine( multiply(n, m) ); } }// This code is contributed by vt_m.

PHP

< ?php // PHP program to find multiplication // of two number without use of // multiplication operator// Function for multiplication function multiply( $n , $m ) { $ans = 0; $count = 0; while ( $m ) { // check for set bit and left // shift n, count times if ( $m % 2 == 1) $ans += $n < < $count ; // increment of place value (count) $count ++; $m /= 2; } return $ans ; }// Driver code $n = 20 ; $m = 13; echo multiply( $n , $m ); // This code is contributed by anuj_67. ?>

输出：