# 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.

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 (&Sum; 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 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



Output :

74970000000035

See this for sample run.

Time Complexity: O(log b)

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