两个数字相同位数的乘积之和

给定两个正整数N1和N2 ，任务是查找两个数字的相同位数的乘积之和。
注意：对于长度不相等的数字，较小数字的前几位需要视为0。
例子：

Input: N1 = 5, N2 = 67 
Output: 35 
Explanation: 
At one’s place, we have digits 5 and 7, their product is 35. At ten’s place we have 6 in N2. As N1 has no digit at ten’s place, 6 will be multiplied with 0, leading to no effect on the sum. Hence, the calculated sum is 35.
Input: N1 = 25, N2 = 1548 
Output: 48 
Explanation: 
Sum = 5 * 8 + 2 * 4 + 0 * 5 + 0 * 1 = 48. 
 

方法：
要解决上述问题，我们需要执行以下步骤：

• 提取两个数字的最右边的数字，并将它们相乘，然后将其乘积相加。
• 现在删除数字。
• 继续重复上述两个步骤，直到其中之一减小为0。然后，打印计算出的总和的最终值。

下面是上述方法的实现：

C++

// C++ program to calculate the
// sum of same placed digits
// of two numbers
#include 
using namespace std;
 
int sumOfProductOfDigits(int n1, int n2)
{
    int sum = 0;
     
    // Loop until one of the numbers
    // have no digits remaining
    while (n1 > 0 && n2 > 0)
    {
        sum += ((n1 % 10) * (n2 % 10));
        n1 /= 10;
        n2 /= 10;
    }
    return sum;
}
 
// Driver Code
int main()
{
    int n1 = 25;
    int n2 = 1548;
 
    cout << sumOfProductOfDigits(n1, n2);
}
 
// This code is contributed by grand_master

Java

// Java program to calculate the
// sum of same placed digits of
// two numbers
 
class GFG {
 
    // Function to find the sum of the
    // products of their corresponding digits
    static int sumOfProductOfDigits(int n1,
                                    int n2)
    {
        int sum = 0;
        // Loop until one of the numbers
        // have no digits remaining
        while (n1 > 0 && n2 > 0) {
            sum += ((n1 % 10) * (n2 % 10));
            n1 /= 10;
            n2 /= 10;
        }
 
        return sum;
    }
 
    // Driver Code
    public static void main(String args[])
    {
 
        int n1 = 25;
        int n2 = 1548;
 
        System.out.println(
            sumOfProductOfDigits(n1, n2));
    }
}

Python3

# Python3 program to calculate the
# sum of same placed digits
# of two numbers
 
def sumOfProductOfDigits(n1, n2):
 
    sum1 = 0;
     
    # Loop until one of the numbers
    # have no digits remaining
    while (n1 > 0 and n2 > 0):
 
        sum1 += ((n1 % 10) * (n2 % 10));
        n1 = n1 // 10;
        n2 = n2 // 10;
         
    return sum1;
 
# Driver Code
n1 = 25;
n2 = 1548;
 
print(sumOfProductOfDigits(n1, n2));
 
# This code is contributed by Nidhi_biet

C#

// C# program to calculate the
// sum of same placed digits of
// two numbers
using System;
class GFG{
 
// Function to find the sum of the
// products of their corresponding digits
static int sumOfProductOfDigits(int n1,
                                int n2)
{
    int sum = 0;
     
    // Loop until one of the numbers
    // have no digits remaining
    while (n1 > 0 && n2 > 0)
    {
        sum += ((n1 % 10) * (n2 % 10));
        n1 /= 10;
        n2 /= 10;
    }
    return sum;
}
 
// Driver Code
public static void Main()
{
    int n1 = 25;
    int n2 = 1548;
 
    Console.WriteLine(
            sumOfProductOfDigits(n1, n2));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript program to calculate the
// sum of same placed digits
// of two numbers
 
function sumOfProductOfDigits(n1, n2)
{
    let sum = 0;
     
    // Loop until one of the numbers
    // have no digits remaining
    while (n1 > 0 && n2 > 0)
    {
        sum += ((n1 % 10) * (n2 % 10));
        n1 = Math.floor(n1/10);
        n2 = Math.floor(n2/10);
    }
    return sum;
}
 
// Driver Code
 
    let n1 = 25;
    let n2 = 1548;
 
    document.write(sumOfProductOfDigits(n1, n2));
 
// This code is contributed by Mayank Tyagi
 

输出：

48