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How to compute mod of a big number?

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  • Difficulty Level : Medium
  • Last Updated : 14 Jun, 2022
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Given a big number ‘num’ represented as string and an integer x, find value of “num % x” or “num mod x”. Output is expected as an integer.

Examples : 

Input:  num = "12316767678678",  a = 10
Output: num (mod a) ≡ 8

The idea is to process all digits one by one and use the property that 

xy (mod a) ≡ ((x (mod a) * 10) + (y (mod a))) mod a

where, x : left-most digit

y: rest of the digits except x.

for example: 

625 % 5 = (((6 % 5)*10) + (25 % 5)) % 5 = 0

Below is the implementation.

Thanks to utkarsh111 for suggesting the below solution.

C++




// C++ program to compute mod of a big number represented
// as string
#include <iostream>
using namespace std;
 
// Function to compute num (mod a)
int mod(string num, int a)
{
    // Initialize result
    int res = 0;
 
    // One by one process all digits of 'num'
    for (int i = 0; i < num.length(); i++)
        res = (res * 10 + (int)num[i] - '0') % a;
 
    return res;
}
 
// Driver program
int main()
{
    string num = "12316767678678";
    cout << mod(num, 10);
    return 0;
}

Java




// Java program to compute mod of a big
// number represented as string
import java.io.*;
 
class GFG {
 
    // Function to compute num (mod a)
    static int mod(String num, int a)
    {
 
        // Initialize result
        int res = 0;
 
        // One by one process all digits of 'num'
        for (int i = 0; i < num.length(); i++)
            res = (res * 10 + (int)num.charAt(i) - '0') % a;
 
        return res;
    }
 
    // Driver program
    public static void main(String[] args)
    {
 
        String num = "12316767678678";
 
        System.out.println(mod(num, 10));
    }
}
 
// This code is contributed by vt_m.

Python3




# program to compute mod of a big number
# represented as string
 
# Function to compute num (mod a)
 
 
def mod(num, a):
 
    # Initialize result
    res = 0
 
    # One by one process all digits
    # of 'num'
    for i in range(0, len(num)):
        res = (res * 10 + int(num[i])) % a
 
    return res
 
 
# Driver program
num = "12316767678678"
print(mod(num, 10))
 
# This code is contributed by Sam007

C#




// C# program to compute mod of a big
// number represented as string
using System;
 
public class GFG {
 
    // Function to compute num (mod a)
    static int mod(String num, int a)
    {
 
        // Initialize result
        int res = 0;
 
        // One by one process all
        // digits of 'num'
        for (int i = 0; i < num.Length; i++)
            res = (res * 10 + (int)num[i] - '0') % a;
 
        return res;
    }
 
    // Driver code
    public static void Main()
    {
        String num = "12316767678678";
 
        Console.WriteLine(mod(num, 10));
    }
}
 
// This code is contributed by Sam007

PHP




<?php
// PHP program to compute mod
// of a big number represented
// as string
 
// Function to compute num (mod a)
function mod($num, $a)
{
    // Initialize result
    $res = 0;
 
    // One by one process
    // all digits of 'num'
    for ($i = 0; $i < $r = strlen($num); $i++)
        $res = ($res * 10 +
                $num[$i] - '0') % $a;
 
    return $res;
}
 
// Driver Code
$num = "12316767678678";
echo mod($num, 10);
 
// This code is contributed by ajit
?>

Javascript




<script>
 
// Javascript program to compute mod
// of a big number represented
// as string
 
// Function to compute num (mod a)
function mod(num, a)
{
     
    // Initialize result
    let res = 0;
 
    // One by one process
    // all digits of 'num'
    for(let i = 0; i < num.length; i++)
        res = (res * 10 +
            parseInt(num[i])) % a;
 
    return res;
}
 
// Driver Code
let num = "12316767678678";
 
document.write(mod(num, 10));
 
// This code is contributed by _saurabh_jaiswal
 
</script>

Output : 

8

Time Complexity : O(|num|)

  • Time complexity will become size of num string as we are traversing once in num.

Auxiliary Space: O(1)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 


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