Program to add two integers of given base
Given three integers X, Y and B, where X and Y are Base-B integers. The task is to find the sum of integers X and Y.
Examples:
Input: X = 123, Y = 234, B = 6
Output: 401
Explanation:
Sum of two integers in base 6 -
1 1
1 2 3
+ 2 3 4
-------------
4 0 1
Input: X = 546, Y = 248 B = 9
Output: 805
Explanation:
Sum of two integers in base 9 -
1 1
5 4 6
+ 2 4 8
-------------
8 0 5
Approach: The idea is to use the fact that whenever two digits of the numbers are added, then the place value will be the modulo of the sum of digits by the base whereas carry will be the integer division of the sum of digits by base. i.e.
Let two digits of the number be D1 and D2 - Place Value = (D1 + D2) % B Carry = (D1 + D2) / B
Similarly, Add every digit from the last to get the desired result.
Below is the implementation of the above approach:
C++
// C++ implementation to find the // sum of two integers of base B #include <bits/stdc++.h> using namespace std; // Function to find the sum of // two integers of base B string sumBaseB(string a, string b, int base) { int len_a, len_b; len_a = a.size(); len_b = b.size(); string sum, s; s = ""; sum = ""; int diff; diff = abs(len_a - len_b); // Padding 0 in front of the // number to make both numbers equal for (int i = 1; i <= diff; i++) s += "0"; // Condition to check if the strings // have lengths mis-match if (len_a < len_b) a = s + a; else b = s + b; int curr, carry = 0; // Loop to find the find the sum // of two integers of base B for (int i = max(len_a, len_b) - 1; i > -1; i--) { // Current Place value for // the resultant sum curr = carry + (a[i] - '0') + (b[i] - '0'); // Update carry carry = curr / base; // Find current digit curr = curr % base; // Update sum result sum = (char)(curr + '0') + sum; } if (carry > 0) sum = (char)(carry + '0') + sum; return sum; } // Driver Code int main() { string a, b, sum; int base; a = "123"; b = "234"; base = 6; // Function Call sum = sumBaseB(a, b, base); cout << sum << endl; return 0; } |
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Java
// Java implementation to find the // sum of two integers of base B class GFG { // Function to find the sum of // two integers of base B static String sumBaseB(String a, String b, int base) { int len_a, len_b; len_a = a.length(); len_b = b.length(); String sum, s; s = ""; sum = ""; int diff; diff = Math.abs(len_a - len_b); // Padding 0 in front of the // number to make both numbers equal for (int i = 1; i <= diff; i++) s += "0"; // Condition to check if the strings // have lengths mis-match if (len_a < len_b) a = s + a; else b = s + b; int curr, carry = 0; // Loop to find the find the sum // of two integers of base B for (int i = Math.max(len_a, len_b) - 1; i > -1; i--) { // Current Place value for // the resultant sum curr = carry + (a.charAt(i) - '0') + (b.charAt(i) - '0'); // Update carry carry = curr / base; // Find current digit curr = curr % base; // Update sum result sum = (char)(curr + '0') + sum; } if (carry > 0) sum = (char)(carry + '0') + sum; return sum; } // Driver Code public static void main (String[] args) { String a, b, sum; int base; a = "123"; b = "234"; base = 6; // Function Call sum = sumBaseB(a, b, base); System.out.println(sum); } } // This code is contributed by AnkitRai01 |
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Python 3
# Python 3 implementation to find the # sum of two integers of base B # Function to find the sum of # two integers of base B def sumBaseB(a,b,base): len_a = len(a) len_b = len(b) s = ""; sum = ""; diff = abs(len_a - len_b); # Padding 0 in front of the # number to make both numbers equal for i in range(1,diff+1): s += "0" # Condition to check if the strings # have lengths mis-match if (len_a < len_b): a = s + a else: b = s + b; carry = 0; # Loop to find the find the sum # of two integers of base B for i in range(max(len_a, len_b) - 1,-1,-1): # Current Place value for # the resultant sum curr = carry + (ord(a[i]) -ord('0')) +( ord(b[i]) - ord('0')); # Update carry carry = curr // base # Find current digit curr = curr % base; # Update sum result sum = chr(curr + ord('0')) + sum if (carry > 0): sum = chr(carry + ord('0')) + sum; return sum # Driver Code a = "123"b = "234"base = 6 # Function Call sum = sumBaseB(a, b, base); print(sum) # This code is contributed by atul_kumar_shrivastava |
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C#
// C# implementation to find the // sum of two integers of base B using System; class GFG { // Function to find the sum of // two integers of base B static string sumBaseB(string a, string b, int base_var) { int len_a, len_b; len_a = a.Length; len_b = b.Length; string sum, s; s = ""; sum = ""; int diff; diff = Math.Abs(len_a - len_b); // Padding 0 in front of the // number to make both numbers equal for (int i = 1; i <= diff; i++) s += "0"; // Condition to check if the strings // have lengths mis-match if (len_a < len_b) a = s + a; else b = s + b; int curr, carry = 0; // Loop to find the find the sum // of two integers of base B for (int i = Math.Max(len_a, len_b) - 1; i > -1; i--) { // Current Place value for // the resultant sum curr = carry + (a[i] - '0') + (b[i] - '0'); // Update carry carry = curr / base_var; // Find current digit curr = curr % base_var; // Update sum result sum = (char)(curr + '0') + sum; } if (carry > 0) sum = (char)(carry + '0') + sum; return sum; } // Driver Code public static void Main (string[] args) { string a, b, sum; int base_var; a = "123"; b = "234"; base_var = 6; // Function Call sum = sumBaseB(a, b, base_var); Console.WriteLine(sum); } } // This code is contributed by AnkitRai01 |
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Output:
401
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