Multiply large integers under large modulo (original) (raw)

Last Updated : 23 Jul, 2025

Given an integer a, b, m. Find (a * b ) mod m, where a, b may be large and their direct multiplication may cause overflow. However, they are smaller than half of the maximum allowed long long int value.

Examples:

Input: a = 426, b = 964, m = 235
Output: 119
Explanation: (426 * 964) % 235 = 410664 % 235 = 119

Input: a = 10123465234878998,
b = 65746311545646431
m = 10005412336548794
Output: 4652135769797794

Naive Approach: A naive approach is to use arbitrary precision data types such as int in python or Biginteger class in Java. But that approach will not be fruitful because the internal conversion of string to int and then perform operation will lead to slow down the calculations of addition and multiplications in the binary number system.

Efficient Approach: Since a and b may be very large numbers, if we try to multiply directly, they will definitely overflow. Therefore we use the basic approach of multiplication i.e., a * b = a + a + ... + a (b times). Now easily compute the value of addition (under modulo m) without any overflow in the calculation. But if we try to add the value of a repeatedly up to b times then it will definitely timeout for the large value of b, since the time complexity of this approach would become O(b).

So, divide the above-repeated steps for a in simpler way i.e.,

If b is even then
a * b = 2 * a * (b / 2),

otherwise
a * b = a + a * (b - 1)

Below is the approach describing the above explanation :

C++ `

// C++ program of finding modulo multiplication #include <bits/stdc++.h>

using namespace std;

// Returns (a * b) % mod long long moduloMultiplication(long long a, long long b, long long mod) { long long res = 0; // Initialize result

// Update a if it is more than
// or equal to mod
a %= mod;

while (b) {
    // If b is odd, add a with result
    if (b & 1)
        res = (res + a) % mod;

    // Here we assume that doing 2*a
    // doesn't cause overflow
    a = (2 * a) % mod;

    b >>= 1; // b = b / 2
}

return res;

}

// Driver program int main() { long long a = 426; long long b = 964; long long m = 235; cout << moduloMultiplication(a, b, m); return 0; }

// This code is contributed // by Akanksha Rai

C

// C program of finding modulo multiplication #include<stdio.h>

// Returns (a * b) % mod long long moduloMultiplication(long long a, long long b, long long mod) { long long res = 0; // Initialize result

// Update a if it is more than
// or equal to mod
a %= mod;

while (b)
{
    // If b is odd, add a with result
    if (b & 1)
        res = (res + a) % mod;

    // Here we assume that doing 2*a
    // doesn't cause overflow
    a = (2 * a) % mod;

    b >>= 1;  // b = b / 2
}

return res;

}

// Driver program int main() { long long a = 10123465234878998; long long b = 65746311545646431; long long m = 10005412336548794; printf("%lld", moduloMultiplication(a, b, m)); return 0; }

Java

// Java program of finding modulo multiplication import java.util.; import java.io.; class GFG {

// Returns (a * b) % mod 
static long moduloMultiplication(long a,
                        long b, long mod)
{
    
    // Initialize result
    long res = 0;  

    // Update a if it is more than 
    // or equal to mod 
    a %= mod;

    while (b > 0) 
    {
        
        // If b is odd, add a with result 
        if ((b & 1) > 0) 
        {
            res = (res + a) % mod;
        }

        // Here we assume that doing 2*a 
        // doesn't cause overflow 
        a = (2 * a) % mod;

        b >>= 1; // b = b / 2 
    }
    return res;
}

// Driver code 
public static void main(String[] args) 
{
    long a = 10123465234878998L;
    long b = 65746311545646431L;
    long m = 10005412336548794L;
    System.out.print(moduloMultiplication(a, b, m));
}

}

// This code is contributed by Rajput-JI

Python3

Python 3 program of finding

modulo multiplication

Returns (a * b) % mod

def moduloMultiplication(a, b, mod):

res = 0; # Initialize result

# Update a if it is more than
# or equal to mod
a = a % mod;

while (b):

    # If b is odd, add a with result
    if (b & 1):
        res = (res + a) % mod;
        
    # Here we assume that doing 2*a
    # doesn't cause overflow
    a = (2 * a) % mod;

    b >>= 1; # b = b / 2

return res;

Driver Code

a = 10123465234878998; b = 65746311545646431; m = 10005412336548794; print(moduloMultiplication(a, b, m));

This code is contributed

by Shivi_Aggarwal

C#

// C# program of finding modulo multiplication using System;

class GFG {

// Returns (a * b) % mod static long moduloMultiplication(long a, long b, long mod) { long res = 0; // Initialize result

// Update a if it is more than
// or equal to mod
a %= mod;

while (b > 0)
{
    // If b is odd, add a with result
    if ((b & 1) > 0)
        res = (res + a) % mod;

    // Here we assume that doing 2*a
    // doesn't cause overflow
    a = (2 * a) % mod;

    b >>= 1; // b = b / 2
}

return res;

}

// Driver code static void Main() { long a = 10123465234878998; long b = 65746311545646431; long m = 10005412336548794; Console.WriteLine(moduloMultiplication(a, b, m)); } }

// This code is contributed // by chandan_jnu

PHP

JavaScript

`

Time complexity: O(log b), A number n has log(n) bits therefore the loop will run log(b) times.
Auxiliary space: O(1)

Note: Above approach will only work if 2 * m can be represented in standard data type otherwise it will lead to overflow.