Modulo de una string binaria grande

Dada una string binaria grande str y un entero K , la tarea es encontrar el valor de str %K .
Ejemplos: 
 

Entrada: str = “1101”, K = 45 
Salida: 13 
decimal(1101) % 45 = 13 % 45 = 13
Entrada: str = “11010101”, K = 112 
Salida: 101 
decimal(11010101) % 112 = 213 % 112 = 101 
 

Enfoque: Se sabe que (str % K) donde str es una string binaria se puede escribir como ((str[n – 1] * 2 ⁰ ) + (str[n – 2] * 2 ¹ ) + … + (str [0] * 2 ^{n – 1} )) % K que a su vez se puede escribir como (((str[n – 1] * 2 ⁰ ) % K) + ((str[n – 2] * 2 ¹ ) % K ) + … + ((str[0] * 2 ^{n – 1} )) % K) % K . Esto se puede usar para encontrar la respuesta requerida sin convertir realmente la string binaria dada a su equivalente decimal.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the value of (str % k)
int getMod(string str, int n, int k)
{
 
    // pwrTwo[i] will store ((2^i) % k)
    int pwrTwo[n];
    pwrTwo[0] = 1 % k;
    for (int i = 1; i < n; i++)
    {
        pwrTwo[i] = pwrTwo[i - 1] * (2 % k);
        pwrTwo[i] %= k;
    }
 
    // To store the result
    int res = 0;
    int i = 0, j = n - 1;
    while (i < n)
    {
 
        // If current bit is 1
        if (str[j] == '1')
        {
 
            // Add the current power of 2
            res += (pwrTwo[i]);
            res %= k;
        }
        i++;
        j--;
    }
    return res;
}
 
// Driver code
int main()
{
    string str = "1101";
    int n = str.length();
    int k = 45;
 
    cout << getMod(str, n, k) << endl;
}
 
// This code is contributed by ashutosh450

// Java implementation of the approach
import java.util.*;
 
class GFG {
 
    // Function to return the value of (str % k)
    static int getMod(String str, int n, int k)
    {
 
        // pwrTwo[i] will store ((2^i) % k)
        int pwrTwo[] = new int[n];
        pwrTwo[0] = 1 % k;
        for (int i = 1; i < n; i++) {
            pwrTwo[i] = pwrTwo[i - 1] * (2 % k);
            pwrTwo[i] %= k;
        }
 
        // To store the result
        int res = 0;
        int i = 0, j = n - 1;
        while (i < n) {
 
            // If current bit is 1
            if (str.charAt(j) == '1') {
 
                // Add the current power of 2
                res += (pwrTwo[i]);
                res %= k;
            }
            i++;
            j--;
        }
 
        return res;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str = "1101";
        int n = str.length();
        int k = 45;
 
        System.out.print(getMod(str, n, k));
    }
}

# Python3 implementation of the approach
 
# Function to return the value of (str % k)
def getMod(_str, n, k) :
 
    # pwrTwo[i] will store ((2^i) % k)
    pwrTwo = [0] * n
    pwrTwo[0] = 1 % k
    for i in range(1, n):
        pwrTwo[i] = pwrTwo[i - 1] * (2 % k)
        pwrTwo[i] %= k
 
    # To store the result
    res = 0
    i = 0
    j = n - 1
    while (i < n) :
 
        # If current bit is 1
        if (_str[j] == '1') :
 
            # Add the current power of 2
            res += (pwrTwo[i])
            res %= k
             
        i += 1
        j -= 1
 
    return res
 
# Driver code
_str = "1101"
n = len(_str)
k = 45
 
print(getMod(_str, n, k))
 
# This code is contributed by
# divyamohan123

// C# implementation of the above approach
using System;
 
class GFG
{
     
    // Function to return the value of (str % k)
    static int getMod(string str, int n, int k)
    {
        int i;
         
        // pwrTwo[i] will store ((2^i) % k)
        int []pwrTwo = new int[n];
         
        pwrTwo[0] = 1 % k;
         
        for (i = 1; i < n; i++)
        {
            pwrTwo[i] = pwrTwo[i - 1] * (2 % k);
            pwrTwo[i] %= k;
        }
 
        // To store the result
        int res = 0;
        i = 0;
        int j = n - 1;
        while (i < n)
        {
 
            // If current bit is 1
            if (str[j] == '1')
            {
 
                // Add the current power of 2
                res += (pwrTwo[i]);
                res %= k;
            }
            i++;
            j--;
        }
        return res;
    }
 
    // Driver code
    public static void Main()
    {
        string str = "1101";
        int n = str.Length;
        int k = 45;
 
        Console.Write(getMod(str, n, k));
    }
}
 
// This code is contributed by AnkitRai01

<script>
 
// Javascript implementation of the approach
 
// Function to return the value of (str % k)
function getMod(str, n, k)
{
 
    // pwrTwo[i] will store ((2^i) % k)
    var pwrTwo = Array(n);
    pwrTwo[0] = 1 % k;
    for (var i = 1; i < n; i++)
    {
        pwrTwo[i] = pwrTwo[i - 1] * (2 % k);
        pwrTwo[i] %= k;
    }
 
    // To store the result
    var res = 0;
    var i = 0, j = n - 1;
    while (i < n)
    {
 
        // If current bit is 1
        if (str[j] == '1')
        {
 
            // Add the current power of 2
            res += (pwrTwo[i]);
            res %= k;
        }
        i++;
        j--;
    }
    return res;
}
 
// Driver code
var str = "1101";
var n = str.length;
var k = 45;
document.write( getMod(str, n, k));
 
 
</script>

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