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Program to find remainder when large number is divided by r

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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.



Last Updated : 06 Apr, 2023
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