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Compute (a*b)%c such that (a%c) * (b%c) can be beyond range

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Given three numbers a, b and c such that a, b and c can be at most 1016. The task is to compute (a*b)%c 
A simple solution of doing ( (a % c) * (b % c) ) % c would not work here. The problem here is that a and b can be large so when we calculate (a % c) * (b % c), it goes beyond the range that long long int can hold, hence overflow occurs. For example, If a = (1012+7), b = (1013+5), c = (1015+3).
Now long long int can hold upto 4 x 1018(approximately) and a*b is much larger than that.
 

Instead of doing direct multiplication we can add find a + a + ……….(b times) and take modulus with c each time we add a so that overflow don’t take place. But this would be inefficient looking at constraint on a, b and c. We have to somehow calculate (? a) % c in optimized manner. 
We can use divide and conquer to calculate it. The main idea is: 
 

  1. If b is even then a*b = (2*a) * (b/2)
  2. If b is odd then a*b = a + (2*a)*((b-1)/2)

Below is the implementation of the algorithm: 
 

C++




// C++ program to Compute (a*b)%c
// such that (a%c) * (b%c) can be
// beyond range
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
 
// returns (a*b)%c
ll mulmod(ll a,ll b,ll c)
{
    // base case if b==0, return 0
    if (b==0)
        return 0;
 
    // Divide the problem into 2 parts
    ll s = mulmod(a, b/2, c);
 
    // If b is odd, return
    // (a+(2*a)*((b-1)/2))%c
    if (b%2==1)
        return (a%c+2*(s%c)) % c;
 
    // If b is odd, return
    // ((2*a)*(b/2))%c
    else
        return (2*(s%c)) % c;
}
 
// Driver code
int main()
{
    ll a = 1000000000007, b = 10000000000005;
    ll c = 1000000000000003;
    printf("%lldn", mulmod(a, b, c));
    return 0;
}


Java




// Java program to Compute (a*b)%c
// such that (a%c) * (b%c) can be
// beyond range
// returns (a*b)%c
class GFG {
 
    static long mulmod(long a, long b, long c) {
        // base case if b==0, return 0
        if (b == 0) {
            return 0;
        }
 
        // Divide the problem into 2 parts
        long s = mulmod(a, b / 2, c);
 
        // If b is odd, return
        // (a+(2*a)*((b-1)/2))%c
        if (b % 2 == 1) {
            return (a % c + 2 * (s % c)) % c;
        } // If b is odd, return
        // ((2*a)*(b/2))%c
        else {
            return (2 * (s % c)) % c;
        }
    }
 
// Driver code
    public static void main(String[] args) {
        long a = 1000000000007L, b = 10000000000005L;
        long c = 1000000000000003L;
        System.out.println((mulmod(a, b, c)));
    }
}
 
// This code is contributed by PrinciRaj1992


Python3




# Python3 program of above approach
 
# returns (a*b)%c
def mulmod(a, b, c):
 
    # base case if b==0, return 0
    if(b == 0):
        return 0
 
    # Divide the problem into 2 parts
    s = mulmod(a, b // 2, c)
 
    # If b is odd, return
    # (a+(2*a)*((b-1)/2))%c
    if(b % 2 == 1):
        return (a % c + 2 * (s % c)) % c
 
    # If b is odd, return
    # ((2*a)*(b/2))%c
    else:
        return (2 * (s % c)) % c
 
# Driver code
if __name__=='__main__':
    a = 1000000000007
    b = 10000000000005
    c = 1000000000000003
    print(mulmod(a, b, c))
 
# This code is contributed by
# Sanjit_Prasad


C#




// C# program to Compute (a*b)%c
// such that (a%c) * (b%c) can be
// beyond range
using System;
 
// returns (a*b)%c
class GFG
{
static long mulmod(long a, long b, long c)
{
    // base case if b==0, return 0
    if (b == 0)
        return 0;
 
    // Divide the problem into 2 parts
    long s = mulmod(a, b / 2, c);
 
    // If b is odd, return
    // (a+(2*a)*((b-1)/2))%c
    if (b % 2 == 1)
        return (a % c + 2 * (s % c)) % c;
 
    // If b is odd, return
    // ((2*a)*(b/2))%c
    else
        return (2 * (s % c)) % c;
}
 
// Driver code
public static void Main()
{
    long a = 1000000000007, b = 10000000000005;
    long c = 1000000000000003;
    Console.WriteLine(mulmod(a, b, c));
}
}
 
// This code is contributed by mits


Javascript




<script>
 
// javascript program to Compute (a*b)%c
// such that (a%c) * (b%c) can be
// beyond range
// returns (a*b)%c  
function mulmod(a , b , c) {
 
        // base case if b==0, return 0
        if (b == 0) {
            return 0;
        }
 
        // Divide the problem into 2 parts
        var s = mulmod(a, parseInt(b / 2), c);
 
        // If b is odd, return
        // (a+(2*a)*((b-1)/2))%c
        if (b % 2 == 1) {
            return (a % c + 2 * (s % c)) % c;
        }
        // If b is odd, return
        // ((2*a)*(b/2))%c
        else {
            return (2 * (s % c)) % c;
        }
    }
 
// Driver code
var a = 1000000000007, b = 10000000000005;
var c = 1000000000000003;
document.write((mulmod(a, b, c)));
 
// This code contributed by Princi Singh
 
</script>


PHP




<?php
// PHP program to Compute (a*b)%c
// such that (a%c) * (b%c) can be
// beyond range
 
// returns (a*b)%c
function mulmod($a, $b, $c)
{
     
    // base case if b==0, return 0
    if ($b==0)
        return 0;
 
    // Divide the problem into 2 parts
    $s = mulmod($a, $b/2, $c);
 
    // If b is odd, return
    // (a+(2*a)*((b-1)/2))%c
    if ($b % 2 == 1)
        return ($a % $c + 2 *
           ($s % $c))  %  $c;
 
    // If b is odd, return
    // ((2*a)*(b/2))%c
    else
        return (2 * ($s % $c)) % $c;
}
 
    // Driver Code
    $a = 1000000000007;
    $b = 10000000000005;
    $c = 1000000000000003;
    echo mulmod($a, $b, $c);
     
// This code is contributed by anuj_67.
?>


Output : 

74970000000035

See this for sample run.
Time Complexity: O(log b)
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Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

Approach#2: Using modulo properties

this  approach is to use the properties of modulo arithmetic. Specifically, we can use the fact that (ab)%c = ((a%c)(b%c))%c. This means that we can first take the modulus of a and b, and then calculate their product.

Algorithm 

Start by defining the function multiply_modulo with three input parameters: a, b, and c.
Calculate the remainder of a when divided by c using the modulo operator and assign it to a.
Calculate the remainder of b when divided by c using the modulo operator and assign it to b.
Multiply a and b.
Calculate the remainder of the product obtained in step 4 when divided by c using the modulo operator and assign it to product.
Return product as the output of the function.

C++




#include <iostream>
#include <sstream>
#include <string>
 
// Function to compute (a*b)%c using modulo properties
std::string multiplyModulo(const std::string& aStr,
                        const std::string& bStr,
                        const std::string& cStr)
{
    // Convert string inputs to BigInteger
    std::stringstream aStream(aStr), bStream(bStr),
        cStream(cStr);
    long long a, b, c;
    aStream >> a;
    bStream >> b;
    cStream >> c;
 
    // Calculate the remainder of a when divided by c
    a = (a % c + c) % c;
 
    // Calculate the remainder of b when divided by c
    b = (b % c + c) % c;
 
    // Multiply a and b
    long long product = (a * b);
 
    // Calculate the remainder of the product when divided
    // by c
    product = (product % c + c) % c;
 
    // Convert the result to string for compatibility with
    // BigInteger
    return std::to_string(product);
}
 
int main()
{
    // Example values
    std::string a = "1000000000007";
    std::string b = "10000000000005";
    std::string c = "1000000000000003";
 
    // Calculate (a*b)%c using the multiplyModulo function
    std::string result = multiplyModulo(a, b, c);
 
    // Print the result
    std::cout << "Result: " << result << std::endl;
 
    return 0;
}


Java




import java.math.BigInteger;
 
public class Main {
    static BigInteger multiplyModulo(BigInteger a, BigInteger b, BigInteger c) {
        a = a.mod(c);
        b = b.mod(c);
        BigInteger product = a.multiply(b).mod(c);
        return product;
    }
 
    public static void main(String[] args) {
        BigInteger a = new BigInteger("1000000000007");
        BigInteger b = new BigInteger("10000000000005");
        BigInteger c = new BigInteger("1000000000000003");
 
        System.out.println(multiplyModulo(a, b, c));
    }
}


Python3




def multiply_modulo(a, b, c):
  a = a % c
  b = b % c
  product = (a * b) % c
  return product
a = 1000000000007
b = 10000000000005
c = 1000000000000003
print(multiply_modulo(a, b, c))


C#




using System;
 
class GFG
{
    // returns (a*b)%c
    static long MultiplyModulo(long a, long b, long c)
    {
        // base case if b==0, return 0
        if (b == 0)
            return 0;
 
        // Divide the problem into 2 parts
        long s = MultiplyModulo(a, b / 2, c);
 
        // If b is odd, return (a+(2*a)*((b-1)/2))%c
        if (b % 2 == 1)
            return (a % c + 2 * (s % c)) % c;
 
        // If b is even, return ((2*a)*(b/2))%c
        else
            return (2 * (s % c)) % c;
    }
 
    // Driver code
    public static void Main()
    {
        long a = 1000000000007, b = 10000000000005;
        long c = 1000000000000003;
        Console.WriteLine(MultiplyModulo(a, b, c));
    }
}


Javascript




// Javascript code addition
 
function multiply_modulo(a, b, c) {
  a = BigInt(a) % BigInt(c);
  b = BigInt(b) % BigInt(c);
  let product = (a * b) % BigInt(c);
  return product;
}
 
let a = BigInt("1000000000007");
let b = BigInt("10000000000005");
let c = BigInt("1000000000000003");
console.log(multiply_modulo(a, b, c).toString());
 
// The code is contributed by Anjali goel.


Output

Result: 972978359410049

Time complexity: O(log(a)+log(b)) 

Auxiliary Space is O(1).



Last Updated : 16 Jan, 2024
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