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
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
C++
#include <bits/stdc++.h>
using namespace std;
int exponentMod(int A, int B, int C)
{
if (A == 0)
return 0;
if (B == 0)
return 1;
long y;
if (B % 2 == 0) {
y = exponentMod(A, B / 2, C);
y = (y * y) % C;
}
else {
y = A % C;
y = (y * exponentMod(A, B - 1, C) % C) % C;
}
return (int)((y + C) % C);
}
int main()
{
int A = 2, B = 5, C = 13;
cout << "Power is " << exponentMod(A, B, C);
return 0;
}
|
C
#include <stdio.h>
int exponentMod(int A, int B, int C)
{
if (A == 0)
return 0;
if (B == 0)
return 1;
long y;
if (B % 2 == 0) {
y = exponentMod(A, B / 2, C);
y = (y * y) % C;
}
else {
y = A % C;
y = (y * exponentMod(A, B - 1, C) % C) % C;
}
return (int)((y + C) % C);
}
int main()
{
int A = 2, B = 5, C = 13;
printf("Power is %d", exponentMod(A, B, C));
return 0;
}
|
Java
import java.io.*;
class GFG
{
static int exponentMod(int A,
int B, int C)
{
if (A == 0)
return 0;
if (B == 0)
return 1;
long y;
if (B % 2 == 0)
{
y = exponentMod(A, B / 2, C);
y = (y * y) % C;
}
else
{
y = A % C;
y = (y * exponentMod(A, B - 1,
C) % C) % C;
}
return (int)((y + C) % C);
}
public static void main(String args[])
{
int A = 2, B = 5, C = 13;
System.out.println("Power is " +
exponentMod(A, B, C));
}
}
|
Python3
def exponentMod(A, B, C):
if (A == 0):
return 0
if (B == 0):
return 1
y = 0
if (B % 2 == 0):
y = exponentMod(A, B / 2, C)
y = (y * y) % C
else:
y = A % C
y = (y * exponentMod(A, B - 1,
C) % C) % C
return ((y + C) % C)
A = 2
B = 5
C = 13
print("Power is", exponentMod(A, B, C))
|
C#
class GFG
{
static int exponentMod(int A, int B, int C)
{
if (A == 0)
return 0;
if (B == 0)
return 1;
long y;
if (B % 2 == 0)
{
y = exponentMod(A, B / 2, C);
y = (y * y) % C;
}
else
{
y = A % C;
y = (y * exponentMod(A, B - 1,
C) % C) % C;
}
return (int)((y + C) % C);
}
public static void Main()
{
int A = 2, B = 5, C = 13;
System.Console.WriteLine("Power is " +
exponentMod(A, B, C));
}
}
|
PHP
<?php
function exponentMod($A, $B, $C)
{
if ($A == 0)
return 0;
if ($B == 0)
return 1;
if ($B % 2 == 0)
{
$y = exponentMod($A, $B / 2, $C);
$y = ($y * $y) % $C;
}
else
{
$y = $A % $C;
$y = ($y * exponentMod($A, $B - 1,
$C) % $C) % $C;
}
return (($y + $C) % $C);
}
$A = 2;
$B = 5;
$C = 13;
echo "Power is " . exponentMod($A, $B, $C);
?>
|
Javascript
<script>
function exponentMod(A, B, C)
{
if (A == 0)
return 0;
if (B == 0)
return 1;
var y;
if (B % 2 == 0)
{
y = exponentMod(A, B / 2, C);
y = (y * y) % C;
}
else
{
y = A % C;
y = (y * exponentMod(A, B - 1,
C) % C) % C;
}
return parseInt(((y + C) % C));
}
var A = 2, B = 5, C = 13;
document.write("Power is " +
exponentMod(A, B, C));
</script>
|
Iterative modular exponentiation.
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