Given two positive integers N1 and N2, the task is to find the sum of the products of the same placed digits of the two numbers.
Note: For numbers of unequal length, the preceding digits of the smaller number needs to be treated as 0.
Examples:
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.
Approach:
To solve the problem mentioned above, we need to follow the steps below:
- Extract the rightmost digits of the two numbers and multiply them and add their product to sum.
- Now remove the digit.
- Keep repeating the above two steps until one of them is reduced to 0. Then, print the final value of sum calculated.
Below is the implementation of the above approach:
C++
// C++ program to calculate the // sum of same placed digits // of two numbers #include <bits/stdc++.h> 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 |
48
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