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:
- If b is even then a*b = (2*a) * (b/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)%cll 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 codeint 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)%cclass 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)%cdef 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 codeif __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 rangeusing System;// returns (a*b)%cclass 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)%cfunction 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)
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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.
Python3
def multiply_modulo(a, b, c): a = a % c b = b % c product = (a * b) % c return producta = 1000000000007b = 10000000000005c = 1000000000000003print(multiply_modulo(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. |
74970000000035
Time complexity: O(log(a)+log(b))
Auxiliary Space is O(1).
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