Given two numnbers **A** and **B**, the task is to perform BCD Addition of the given numbers.

**Examples:**

Input:A = 12, B = 20

Output:110010

Explanation:

The summation of A and B is 12 + 20 = 32.

The binary representation of 3 = 0011

The binary representation of 2 = 0010

Therefore, the BCD Addition is “0011” + “0010” = “110010”

Input:A = 10, B = 10

Output:100000

Explanation:

The summation of A and B is 10 + 10 = 20.

The binary representation of 2 = 0010

The binary representation of 0 = 0000

Therefore, the BCD Addition is “0010” + “0000” = “100000”

**Approach:** The idea is to convert the summation of given two numbers A and B to BCD Number. Below are the steps:

- Find the summation(say
**num**) of the two given numbers**A**and**B**. - For each digit in the number
**num**, convert it into binary representation upto**4 bits**. - Concatenate the binary representation of each digit above and print the result.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to perform BCD Addition ` `string BCDAddition(` `int` `A, ` `int` `B) ` `{ ` ` ` ` ` `// Store the summation of A and B ` ` ` `// in form of string ` ` ` `string s = to_string(A + B); ` ` ` `int` `l = s.length(); ` ` ` ` ` `// To store the final result ` ` ` `string ans; ` ` ` ` ` `string str; ` ` ` ` ` `// Forming BCD using Bitset ` ` ` `for` `(` `int` `i = 0; i < l; i++) { ` ` ` ` ` `// Find the binary representation ` ` ` `// of the current characters ` ` ` `str = bitset<4>(s[i]).to_string(); ` ` ` `ans.append(str); ` ` ` `} ` ` ` ` ` `// Stripping off leading zeroes. ` ` ` `const` `auto` `loc1 = ans.find(` `'1'` `); ` ` ` ` ` `// Return string ans ` ` ` `if` `(loc1 != string::npos) { ` ` ` `return` `ans.substr(loc1); ` ` ` `} ` ` ` `return` `"0"` `; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `// Given Numbers ` ` ` `int` `A = 12, B = 20; ` ` ` ` ` `// Function Call ` ` ` `cout << BCDAddition(A, B); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

110010

**Time Complexity:** *O(log _{10}(A+B))*

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