Programa para sumar dos enteros de base dada

Dados tres enteros X , Y y B , donde X e Y son enteros de base B. La tarea es encontrar la suma de los enteros X e Y.

Ejemplos: 

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   

Enfoque: La idea es utilizar el hecho de que cada vez que se suman dos dígitos de los números, el valor posicional será el módulo de la suma de los dígitos por la base, mientras que el acarreo será la división entera de la suma de los dígitos por la base. es decir  

Let two digits of the number be D1 and D2 -
Place Value = (D1 + D2) % B

Carry = (D1 + D2) / B

Del mismo modo, sume todos los dígitos desde el último para obtener el resultado deseado.

A continuación se muestra la implementación del enfoque anterior: 

// 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;
}

// 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

# 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

// 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

<script>
 
// Javascript implementation to find the
// sum of two integers of base B
 
// Function to find the sum of
// two integers of base B
function sumBaseB(a, b, base_var)
{
    let len_a, len_b;
   
    len_a = a.length;
    len_b = b.length;
   
    let sum, s;
    s = "";
    sum = "";
   
    let diff;
    diff = Math.abs(len_a - len_b);
       
    // Padding 0 in front of the
    // number to make both numbers equal
    for(let 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;
   
    let curr, carry = 0;
       
    // Loop to find the find the sum
    // of two integers of base B
    for(let i = Math.max(len_a, len_b) - 1;
            i > -1; i--)
    {
         
        // Current Place value for
        // the resultant sum
        curr = carry + (a[i].charCodeAt() -
                         '0'.charCodeAt()) +
                       (b[i].charCodeAt() -
                         '0'.charCodeAt());
   
        // Update carry
        carry = parseInt(curr / base_var, 10);
   
        // Find current digit
        curr = curr % base_var;
   
        // Update sum result
        sum = String.fromCharCode(
            curr + '0'.charCodeAt()) + sum;
    }
    if (carry > 0)
        sum = String.fromCharCode(
            carry + '0'.charCodeAt()) + sum;
             
    return sum;
}
 
// Driver code
let a, b, sum;
let base_var;
a = "123";
b = "234";
base_var = 6;
 
// Function Call
sum = sumBaseB(a, b, base_var);
document.write(sum + "</br>");
 
// This code is contributed by divyesh072019  
 
</script>

401

Complejidad de tiempo: O(len_a – len_b)

Espacio Auxiliar: O(1)