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

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++

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// 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;
}

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Java

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// 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 

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Python3

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# 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

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C#

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// 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

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PHP

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<?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.
?>

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Output :

74970000000035

See this for sample run.

Time Complexity: O(log b)

This article is contributed by Madhur Modi. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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