Consider below simple method to multiply two numbers.
// A Simple solution that causes overflow when // value of (a % mod) * (b % mod) becomes more than// maximum value of long long int #define ll long long ll multiply(ll a, ll b, ll mod){ return ((a % mod) * (b % mod)) % mod;} |
The above function works fine when multiplication doesn’t result in overflow. But if input numbers are such that the result of multiplication is more than maximum limit.
For example, the above method fails when mod = 1011, a = 9223372036854775807 (largest long long int) and b = 9223372036854775807 (largest long long int). Note that there can be smaller values for which it may fail. There can be many more examples of smaller values. In fact any set of values for which multiplication can cause a value greater than maximum limit.
How to avoid overflow?
We can multiply recursively to overcome the difficulty of overflow. To multiply a*b, first calculate a*b/2 then add it twice. For calculating a*b/2 calculate a*b/4 and so on (similar to log n exponentiation algorithm).
// To compute (a * b) % mod
multiply(a, b, mod)
1) ll res = 0; // Initialize result
2) a = a % mod.
3) While (b > 0)
a) If b is odd, then add 'a' to result.
res = (res + a) % mod
b) Multiply 'a' with 2
a = (a * 2) % mod
c) Divide 'b' by 2
b = b/2
4) Return res Below is the implementation.
C++
// C++ program for modular multiplication without// any overflow#include<iostream>using namespace std; typedef long long int ll; // To compute (a * b) % modll mulmod(ll a, ll b, ll mod){ ll res = 0; // Initialize result a = a % mod; while (b > 0) { // If b is odd, add 'a' to result if (b % 2 == 1) res = (res + a) % mod; // Multiply 'a' with 2 a = (a * 2) % mod; // Divide b by 2 b /= 2; } // Return result return res % mod;} // Driver programint main(){ ll a = 9223372036854775807, b = 9223372036854775807; cout << mulmod(a, b, 100000000000); return 0;} |
Java
// Java program for modular multiplication // without any overflow class GFG { // To compute (a * b) % mod static long mulmod(long a, long b, long mod) { long res = 0; // Initialize result a = a % mod; while (b > 0) { // If b is odd, add 'a' to result if (b % 2 == 1) { res = (res + a) % mod; } // Multiply 'a' with 2 a = (a * 2) % mod; // Divide b by 2 b /= 2; } // Return result return res % mod; } // Driver code public static void main(String[] args) { long a = 9223372036854775807L, b = 9223372036854775807L; System.out.println(mulmod(a, b, 100000000000L)); }} // This code is contributed by Rajput-JI |
Python3
# Python3 program for modular multiplication # without any overflow # To compute (a * b) % moddef mulmod(a, b, mod): res = 0; # Initialize result a = a % mod; while (b > 0): # If b is odd, add 'a' to result if (b % 2 == 1): res = (res + a) % mod; # Multiply 'a' with 2 a = (a * 2) % mod; # Divide b by 2 b //= 2; # Return result return res % mod; # Driver Codea = 9223372036854775807;b = 9223372036854775807;print(mulmod(a, b, 100000000000)); # This code is contributed by mits |
C#
// C# program for modular multiplication // without any overflow using System; class GFG { // To compute (a * b) % mod static long mulmod(long a, long b, long mod) { long res = 0; // Initialize result a = a % mod; while (b > 0) { // If b is odd, add 'a' to result if (b % 2 == 1) { res = (res + a) % mod; } // Multiply 'a' with 2 a = (a * 2) % mod; // Divide b by 2 b /= 2; } // Return result return res % mod; } // Driver code public static void Main(String[] args) { long a = 9223372036854775807L, b = 9223372036854775807L; Console.WriteLine(mulmod(a, b, 100000000000L)); } } // This code is contributed by 29AjayKumar |
Output:
84232501249
Thanks to Utkarsh Trivedi for suggesting above solution.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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