用移位运算符将两个数相乘

对于任何给定的两个数字n和m，您都必须找到n * m而不使用任何乘法运算符。
例子 ：

Input: n = 25 , m = 13
Output: 325

Input: n = 50 , m = 16
Output: 800

我们可以用shift运算符解决这个问题。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。并且与数字的乘积等效于与2的乘方的乘积。可以使用左移运算符获得2的乘方。
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，如果m的设置位的值置位，则将该值相加并回答。

C++

// CPP program to find multiplication
// of two number without use of
// multiplication operator
#include
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#

// 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.

Javascript

 
// JavaScript program to find multiplication
// of two number without use of
// multiplication operator
 
// Function for multiplication
function multiply(n, m)
{
    let 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 = Math.floor(m / 2);
    }
    return ans;
}
 
// Driver code
 
    let n = 20 , m = 13;
    document.write(multiply(n, m));
  
// This code is contributed by Surbhi Tyagi.
 

输出 ：

260