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

  • Difficulty Level : Hard
  • Last Updated : 19 Apr, 2021

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: 
 

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  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

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.
?>

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>

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




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