Program to find remainder when large number is divided by r

Given a number N, the task is to find the remainder when N is divided by R (a two-digit Number). The input of the Number may be very large.

Examples:

Input: N = 13589234356546756, R = 13
Output: 11

Input: N = 3435346456547566345436457867978, R = 17
Output: 13

• Get the digit of N one by one from left to right.
• For each digit, combine with next digit if its less than R.
• If the combination at any point reaches above R, take and store the Remainder.
• Repeat the above steps for all digits from left to right.

Below is the program that implements the above approach: 

C

// C implementation to find Remainder
// when a large Number is divided by R
#include <stdio.h>
#include <string.h>

// Function to Return Remainder
int Remainder(char str[], int R)
{
    // len is variable to store the
    // length of Number string.
    int len = strlen(str);

    int Num, Rem = 0;

    // loop that find Remainder
    for (int i = 0; i < len; i++) {

        Num = Rem * 10 + (str[i] - '0');
        Rem = Num % R;
    }

    // Return the remainder
    return Rem;
}

// Driver code
int main()
{

    // Get the large number as string
    char str[] = "13589234356546756";

    // Get the divisor R
    int R = 13;

    // Find and print the remainder
    printf("%d",Remainder(str, R));

    return 0;
}

// This code is contributed by kothavvsaakash.

C++

// CPP implementation to find Remainder
// when a large Number is divided by R

#include <bits/stdc++.h>
using namespace std;

// Function to Return Remainder
int Remainder(string str, int R)
{
    // len is variable to store the
    // length of Number string.
    int len = str.length();

    int Num, Rem = 0;

    // loop that find Remainder
    for (int i = 0; i < len; i++) {

        Num = Rem * 10 + (str[i] - '0');
        Rem = Num % R;
    }

    // Return the remainder
    return Rem;
}

// Driver code
int main()
{

    // Get the large number as string
    string str = "13589234356546756";

    // Get the divisor R
    int R = 13;

    // Find and print the remainder
    cout << Remainder(str, R);

    return 0;
}

Java

// Java implementation to find Remainder
// when a large Number is divided by R

class GFG
{
    // Function to Return Remainder
    static int Remainder(String str, int R)
    {
        // len is variable to store the
        // length of Number string.
        int len = str.length();

        int Num, Rem = 0;

        // loop that find Remainder
        for (int i = 0; i < len; i++) {

            Num = Rem * 10 + (str.charAt(i) - '0');
            Rem = Num % R;
        }

        // Return the remainder
        return Rem;
    }

    // Driver code
    public static void main( String [] args)
    {

        // Get the large number as string
        String str = "13589234356546756";

        // Get the divisor R
        int R = 13;

        // Find and print the remainder
        System.out.println(Remainder(str, R));


    }
}

// This code is contributed
// by ihritik

Python 3

# Python 3 implementation to
# find Remainder when a large
# Number is divided by R

# Function to Return Remainder
def Remainder(str, R):

    # len is variable to store the
    # length of Number string.
    l = len(str)

    Rem = 0

    # loop that find Remainder
    for i in range(l):

        Num = Rem * 10 + (ord(str[i]) -
                          ord('0'))
        Rem = Num % R

    # Return the remainder
    return Rem

# Driver code
if __name__ == "__main__":

    # Get the large number
    # as string
    str = "13589234356546756"

    # Get the divisor R
    R = 13

    # Find and print the remainder
    print(Remainder(str, R))

# This code is contributed
# by ChitraNayal

Javascript

<script>

// Javascript implementation to find Remainder
// when a large Number is divided by R

// Function to Return Remainder
function Remainder(str, R)
{
    // len is variable to store the
    // length of Number string.
    var len = str.length;

    var Num, Rem = 0;

    // loop that find Remainder
    for (var i = 0; i < len; i++) {

        Num = Rem * 10 + (str[i] - '0');
        Rem = Num % R;
    }

    // Return the remainder
    return Rem;
}

// Driver code
// Get the large number as string
var str = "13589234356546756";
// Get the divisor R
var R = 13;
// Find and print the remainder
document.write(Remainder(str, R));

// This code is contributed by noob2000.
</script>

C#

// C# implementation to find
// Remainder when a large
// Number is divided by R
using System;

class GFG
{

// Function to Return Remainder
static int Remainder(String str,
                     int R)
{
    // len is variable to store the
    // length of Number string.
    int len = str.Length;

    int Num, Rem = 0;

    // loop that find Remainder
    for (int i = 0; i < len; i++)
    {
        Num = Rem * 10 + (str[i] - '0');
        Rem = Num % R;
    }

    // Return the remainder
    return Rem;
}

// Driver code
public static void Main()
{

    // Get the large number as string
    String str = "13589234356546756";

    // Get the divisor R
    int R = 13;

    // Find and print the remainder
    Console.WriteLine(Remainder(str, R));
}
}

// This code is contributed
// by Subhadeep

PHP

<?php
// PHP implementation to find Remainder
// when a large Number is divided by R

// Function to Return Remainder
function Remainder($str, $R)
{
    // len is variable to store the
    // length of Number string.
    $len = strlen($str);

    $Num = 0; $Rem = 0;

    // loop that find Remainder
    for ($i = 0; $i < $len; $i++)
    {

        $Num = $Rem * 10 + ($str[$i] - '0');
        $Rem = $Num % $R;
    }

    // Return the remainder
    return $Rem;
}

// Driver code

// Get the large number as string
$str = "13589234356546756";

// Get the divisor R
$R = 13;

// Find and print the remainder
echo Remainder($str, $R);

// This code is contributed
// by Akanksha Rai(Abby_akku)

Output

11

Complexity Analysis:

• Time Complexity: O(L) where L is the length of the string
• Auxiliary Space: O(1), since no extra space has been taken.

Another approach:

Approach

1. Define the large number as a string.
2. Define the value of r.
3. Initialize the remainder to 0.
4. Iterate over each digit of the number using a loop.
5. Convert the current digit from a character to an integer.
6. Update the remainder by applying the modulo operator to the digit and the current remainder.
7. Print the final remainder.

C++

#include <iostream>

int main()
{
    // Define the large number as a string
    std::string number
        = "123456789123456789123456789123456789123456789";
    // Define the value of r
    int r = 7;

    // Initialize the remainder to 0
    int remainder = 0;

    // Iterate over each digit of the number
    for (int i = 0; i < number.length(); i++) {
        // Convert the digit from a character to an integer
        int digit = number[i] - '0';

        // Update the remainder by applying the modulo
        // operator to the digit and the current remainder
        remainder = (remainder * 10 + digit) % r;
    }

    // Print the remainder
    std::cout << "The remainder when " << number
              << " is divided by " << r << " is "
              << remainder << "." << std::endl;

    return 0;
}

C

#include <stdio.h>

int main() {
    // Define the large number as a string
    char number[] = "123456789123456789123456789123456789123456789";

    // Define the value of r
    int r = 7;

    // Initialize the remainder to 0
    int remainder = 0;

    // Iterate over each digit of the number
    for (int i = 0; number[i] != '\0'; i++) {
        // Convert the digit from a character to an integer
        int digit = number[i] - '0';

        // Update the remainder by applying the modulo operator to the digit and the current remainder
        remainder = (remainder * 10 + digit) % r;
    }

    // Print the remainder
    printf("The remainder when %s is divided by %d is %d.\n", number, r, remainder);

    return 0;
}

Python3

# Define the large number as a string
number = "123456789123456789123456789123456789123456789"
# Define the value of r
r = 7

# Initialize the remainder to 0
remainder = 0

# Iterate over each digit of the number
for i in range(len(number)):
    # Convert the digit from a character to an integer
    digit = int(number[i])

    # Update the remainder by applying the modulo
    # operator to the digit and the current remainder
    remainder = (remainder * 10 + digit) % r

# Print the remainder
print(f"The remainder when {number} is divided by {r} is {remainder}.")

Javascript

// Define the large number as a string
let number = "123456789123456789123456789123456789123456789";
// Define the value of r
let r = 7;

// Initialize the remainder to 0
let remainder = 0;

// Iterate over each digit of the number
for (let i = 0; i < number.length; i++) {
    // Convert the digit from a character to an integer
    let digit = parseInt(number.charAt(i));

    // Update the remainder by applying the modulo
    // operator to the digit and the current remainder
    remainder = (remainder * 10 + digit) % r;
}

// Print the remainder
console.log(`The remainder when ${number} is divided by ${r} is ${remainder}.`);

Java

import java.math.BigInteger;

public class Main {
    public static void main(String[] args)
    {
        // Define the large number as a string
        String number
            = "123456789123456789123456789123456789123456789";
        // Define the value of r
        int r = 7;
        // Convert the string to a BigInteger
        BigInteger bigNumber = new BigInteger(number);

        // Compute the remainder when the BigInteger is
        // divided by r
        BigInteger remainder
            = bigNumber.remainder(BigInteger.valueOf(r));

        // Print the remainder
        System.out.println("The remainder when " + number
                           + " is divided by " + r + " is "
                           + remainder + ".");
    }
}

C#

using System;

class Program
{
    static void Main()
    {
        // Define the large number as a string
        string number = "123456789123456789123456789123456789123456789";
        // Define the value of r
        int r = 7;

        // Initialize the remainder to 0
        int remainder = 0;

        // Iterate over each digit of the number
        for (int i = 0; i < number.Length; i++)
        {
            // Convert the digit from a character to an integer
            int digit = number[i] - '0';

            // Update the remainder by applying the modulo
            // operator to the digit and the current remainder
            remainder = (remainder * 10 + digit) % r;
        }

        // Print the remainder
        Console.WriteLine("The remainder when {0} is divided by {1} is {2}.", number, r, remainder);
    }
}

Output

The remainder when 123456789123456789123456789123456789123456789 is divided by 7 is 1.

Time complexity: O(n), where n is the number of digits in the large number.
Auxiliary Space: O(1), as we are only storing a few integers and a string of digits.