Given three numbers a, b and c, we need to find (ab) % c
Now why do “% c” after exponentiation, because ab will be really large even for relatively small values of a, b and that is a problem because the data type of the language that we try to code the problem, will most probably not let us store such a large number.
Examples:
Input : a = 2312 b = 3434 c = 6789 Output : 6343 Input : a = -3 b = 5 c = 89 Output : 24
Auxiliary Space: O(1)
The idea is based on below properties.
Property 1:
(m * n) % p has a very interesting property:
(m * n) % p =((m % p) * (n % p)) % p
Property 2:
if b is even:
(a ^ b) % c = ((a ^ b/2) * (a ^ b/2))%c ? this suggests divide and conquer
if b is odd:
(a ^ b) % c = (a * (a ^( b-1))%c
Property 3:
If we have to return the mod of a negative number x whose absolute value is less than y:
then (x + y) % y will do the trick
Note:
Also as the product of (a ^ b/2) * (a ^ b/2) and a * (a ^( b-1) may cause overflow, hence we must be careful about those scenarios
// Recursive C++ program to compute modular power #include <bits/stdc++.h> using namespace std;
int exponentMod( int A, int B, int C)
{ // Base cases
if (A == 0)
return 0;
if (B == 0)
return 1;
// If B is even
long y;
if (B % 2 == 0) {
y = exponentMod(A, B / 2, C);
y = (y * y) % C;
}
// If B is odd
else {
y = A % C;
y = (y * exponentMod(A, B - 1, C) % C) % C;
}
return ( int )((y + C) % C);
} // Driver code int main()
{ int A = 2, B = 5, C = 13;
cout << "Power is " << exponentMod(A, B, C);
return 0;
} // This code is contributed by SHUBHAMSINGH10 |
// Recursive C program to compute modular power #include <stdio.h> int exponentMod( int A, int B, int C)
{ // Base cases
if (A == 0)
return 0;
if (B == 0)
return 1;
// If B is even
long y;
if (B % 2 == 0) {
y = exponentMod(A, B / 2, C);
y = (y * y) % C;
}
// If B is odd
else {
y = A % C;
y = (y * exponentMod(A, B - 1, C) % C) % C;
}
return ( int )((y + C) % C);
} // Driver program to test above functions int main()
{ int A = 2, B = 5, C = 13;
printf ( "Power is %d" , exponentMod(A, B, C));
return 0;
} |
// Recursive Java program // to compute modular power import java.io.*;
class GFG
{ static int exponentMod( int A,
int B, int C)
{ // Base cases
if (A == 0 )
return 0 ;
if (B == 0 )
return 1 ;
// If B is even
long y;
if (B % 2 == 0 )
{
y = exponentMod(A, B / 2 , C);
y = (y * y) % C;
}
// If B is odd
else
{
y = A % C;
y = (y * exponentMod(A, B - 1 ,
C) % C) % C;
}
return ( int )((y + C) % C);
} // Driver Code public static void main(String args[])
{ int A = 2 , B = 5 , C = 13 ;
System.out.println( "Power is " +
exponentMod(A, B, C));
} } // This code is contributed // by Swetank Modi. |
# Recursive Python program # to compute modular power def exponentMod(A, B, C):
# Base Cases
if (A = = 0 ):
return 0
if (B = = 0 ):
return 1
# If B is Even
y = 0
if (B % 2 = = 0 ):
y = exponentMod(A, B / 2 , C)
y = (y * y) % C
# If B is Odd
else :
y = A % C
y = (y * exponentMod(A, B - 1 ,
C) % C) % C
return ((y + C) % C)
# Driver Code A = 2
B = 5
C = 13
print ( "Power is" , exponentMod(A, B, C))
# This code is contributed # by Swetank Modi. |
// Recursive C# program // to compute modular power class GFG
{ static int exponentMod( int A, int B, int C)
{ // Base cases
if (A == 0)
return 0;
if (B == 0)
return 1;
// If B is even
long y;
if (B % 2 == 0)
{
y = exponentMod(A, B / 2, C);
y = (y * y) % C;
}
// If B is odd
else
{
y = A % C;
y = (y * exponentMod(A, B - 1,
C) % C) % C;
}
return ( int )((y + C) % C);
} // Driver Code public static void Main()
{ int A = 2, B = 5, C = 13;
System.Console.WriteLine( "Power is " +
exponentMod(A, B, C));
} } // This code is contributed // by mits |
<?php // Recursive PHP program to // compute modular power function exponentMod( $A , $B , $C )
{ // Base cases
if ( $A == 0)
return 0;
if ( $B == 0)
return 1;
// If B is even
if ( $B % 2 == 0)
{
$y = exponentMod( $A , $B / 2, $C );
$y = ( $y * $y ) % $C ;
}
// If B is odd
else
{
$y = $A % $C ;
$y = ( $y * exponentMod( $A , $B - 1,
$C ) % $C ) % $C ;
}
return (( $y + $C ) % $C );
} // Driver Code $A = 2;
$B = 5;
$C = 13;
echo "Power is " . exponentMod( $A , $B , $C );
// This code is contributed // by Swetank Modi. ?> |
<script> // Recursive Javascript program // to compute modular power // Function to check if a given // quadrilateral is valid or not function exponentMod(A, B, C)
{ // Base cases
if (A == 0)
return 0;
if (B == 0)
return 1;
// If B is even
var y;
if (B % 2 == 0)
{
y = exponentMod(A, B / 2, C);
y = (y * y) % C;
}
// If B is odd
else
{
y = A % C;
y = (y * exponentMod(A, B - 1,
C) % C) % C;
}
return parseInt(((y + C) % C));
} // Driver code var A = 2, B = 5, C = 13;
document.write( "Power is " +
exponentMod(A, B, C));
// This code is contributed by Khushboogoyal499 </script> |
Power is 6
Time Complexity : O(logn)
Auxiliary Space: O(logn)
Iterative modular exponentiation.