Addition of two numbers without carry

You are given two positive number n and m. You have to find simply addition of both number but with a given condition that there is not any carry system in this addition. That is no carry is added at higher MSBs.

Examples :

Input : m = 456, n = 854
Output : 200

Input : m = 456, n = 4
Output : 450

Algorithm :

• Input n, m
• while(n||m) { // Add each bits bit_sum = (n%10) + (m%10); // Neglect carry bit_sum %= 10; // Update result // multiplier to maintain place value res = (bit_sum * multiplier) + res; n /= 10; m /= 10; // Update multiplier multiplier *=10; }
• print res
• Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach :
To solve this problem we will need the bit by bit addition of number where we start adding two number from right most bit (LSB) and add integers from both nubers with same position. Also we will neglect carry at each position so that that carry will not affect further higher bit position.
Start adding both numbers bit by bit and for each bit take sum of integers then neglect their carry by taking modulo of bit_sum by 10 further add bit_sum to res by multiplying bit_sum with a multiplier specifying place value. (Multiplier got incremented 10 times on each iteration.)

Below is the implementation of above approach :

C++

 // CPP program for special  // addition of two number #include using namespace std;    int xSum(int n, int m) {     // variable to store result        int res = 0;        // variable to maintain      // place value     int multiplier = 1;        // variable to maintain      // each digit sum     int bit_sum;        // Add numbers till each      // number become zero     while (n || m) {            // Add each bits         bit_sum = (n % 10) + (m % 10);                    // Neglect carry         bit_sum %= 10;                    // Update result         res = (bit_sum * multiplier) + res;         n /= 10;         m /= 10;                    // Update multiplier         multiplier *= 10;     }     return res; }    // Driver program int main() {     int n = 8458;     int m = 8732;     cout << xSum(n, m);     return 0; }

Java

 // Java program for special  // addition of two number import java.util.*; import java.lang.*;    public class GfG {        public static int xSum(int n, int m)     {         int res = 0;         int multiplier = 1;         int bit_sum;            // Add numbers till each          // number become zero         while (true) {                                if(n==0 && m==0)             break;                // Add each bits             bit_sum = (n % 10) + (m % 10);                // Neglect carry             bit_sum %= 10;                // Update result             res = (bit_sum * multiplier) + res;             n /= 10;             m /= 10;                // Update multiplier             multiplier *= 10;                      }         return res;     }        // Driver function     public static void main(String args[])     {         int n = 8458;         int m = 8732;         System.out.println(xSum(n, m));     } } /* This code is contributed by Sagar Shukla */

Python3

 # Python3 program for special  # addition of two number import math    def xSum(n, m) :        # variable to      # store result      res = 0        # variable to maintain      # place value     multiplier = 1        # variable to maintain      # each digit sum     bit_sum = 0        # Add numbers till each      # number become zero     while (n or m) :            # Add each bits         bit_sum = ((n % 10) +                     (m % 10))                    # Neglect carry         bit_sum = bit_sum % 10                    # Update result         res = (bit_sum *                multiplier) + res         n = math.floor(n / 10)         m = math.floor(m / 10)                    # Update multiplier         multiplier = multiplier * 10            return res    # Driver code n = 8458 m = 8732 print (xSum(n, m))    # This code is contributed by # Manish Shaw(manishshaw1)

C#

 // C# program for special  // addition of two number using System;    public class GfG {        public static int xSum(int n, int m)     {         int res = 0;         int multiplier = 1;         int bit_sum;            // Add numbers till each          // number become zero         while (true) {                // Add each bits             bit_sum = (n % 10) + (m % 10);                // Neglect carry             bit_sum %= 10;                // Update result             res = (bit_sum * multiplier) + res;             n /= 10;             m /= 10;                // Update multiplier             multiplier *= 10;             if (n == 0)                 break;             if (m == 0)                 break;         }         return res;     }        // Driver function     public static void Main()     {         int n = 8458;         int m = 8732;         Console.WriteLine(xSum(n, m));     } }    /* This code is contributed by Vt_m */

PHP



Output :

6180

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