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 :
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 <bits/stdc++.h> 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 = 8458m = 8732print (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
<?php // php program for special // addition of two number function xSum($n, $m) { // variable to store result $res = 0; // variable to maintain // place value $multiplier = 1; // variable to maintain // each digit sum $bit_sum; // Add numbers till each // number become zero while ($n || $m) { // Add each bits $bit_sum = ($n % 10) + ($m % 10); // Neglect carry $bit_sum %= 10; // Update result $res = ($bit_sum * $multiplier) + $res; $n =floor($n / 10); $m =floor($m / 10); // Update multiplier $multiplier *= 10; } return $res; } // Driver code $n = 8458; $m = 8732; echo xSum($n, $m); //This code is contributed by mits ?> |
Output :
6180
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