Given x, k and m. Compute (xxxx…k)%m, x is in power k times. Given x is always prime and m is greater than x.
Examples:
Input : 2 3 3 Output : 1 Explanation : ((2 ^ 2) ^ 2) % 3 = (4 ^ 2) % 3 = 1 Input : 3 2 3 Output : 0 Explanation : (3^3)%3 = 0
A naive approach is to compute the power of x k times and do modulus operation every time.
// C++ program for computing // x^x^x^x.. %m #include <bits/stdc++.h> using namespace std;
// Function to compute the given value int calculate( int x, int k, int m)
{ int result = x;
k--;
// compute power k times
while (k--) {
result = pow (result, x);
if (result > m)
result %= m;
}
return result;
} // Driver Code int main()
{ int x = 5, k = 2, m = 3;
// Calling function
cout << calculate(x, k, m);
return 0;
} |
// C program for computing // x^x^x^x.. %m #include <stdio.h> #include <math.h> // Function to compute the given value int calculate( int x, int k, int m)
{ int result = x;
k--;
// compute power k times
while (k--) {
result = pow (result, x);
if (result > m)
result %= m;
}
return result;
} // Driver Code int main()
{ int x = 5, k = 2, m = 3;
// Calling function
printf ( "%d" ,calculate(x, k, m));
return 0;
} // This code is contributed by kothavvsaakash. |
// Java program for computing // x^x^x^x.. %m class GFG
{ // Function to compute // the given value static int calculate( int x,
int k, int m)
{ int result = x;
k--;
// compute power k times
while (k --> 0 )
{
result = ( int )Math.pow(result, x);
if (result > m)
result %= m;
}
return result;
} // Driver Code public static void main(String args[])
{ int x = 5 , k = 2 , m = 3 ;
// Calling function
System.out.println( calculate(x, k, m));
} } // This code is contributed by Arnab Kundu |
# Python3 program for # computing x^x^x^x.. %m import math
# Function to compute # the given value def calculate(x, k, m):
result = x;
k = k - 1 ;
# compute power k times
while (k):
result = math. pow (result, x);
if (result > m):
result = result % m;
k = k - 1 ;
return int (result);
# Driver Code x = 5 ;
k = 2 ;
m = 3 ;
# Calling function print (calculate(x, k, m));
# This code is contributed # by mits |
// C# program for computing // x^x^x^x.. %m using System;
class GFG
{ // Function to compute // the given value static int calculate( int x,
int k,
int m)
{ int result = x;
k--;
// compute power
// k times
while (k --> 0)
{
result = ( int )Math.Pow(result, x);
if (result > m)
result %= m;
}
return result;
} // Driver Code static public void Main ()
{ int x = 5, k = 2, m = 3;
// Calling function
Console.WriteLine(
calculate(x, k, m));
} } // This code is contributed // by ajit |
<?php // PHP program for computing // x^x^x^x.. %m // Function to compute // the given value function calculate( $x , $k , $m )
{ $result = $x ;
$k --;
// compute power k times
while ( $k --)
{
$result = pow( $result , $x );
if ( $result > $m )
$result %= $m ;
}
return $result ;
} // Driver Code $x = 5;
$k = 2;
$m = 3;
// Calling function echo calculate( $x , $k , $m );
// This code is contributed // by akt_mit ?> |
<script> //program for computing // x^x^x^x.. %m // Function to compute // the given value function calculate(x, k, m)
{ let result = x;
k = k - 1;
// compute power k times
while (k--)
{
result = Math.pow(result, x);
if (result > m)
result %= m;
}
return result;
} // Driver Code let x = 5; let k = 2; let m = 3; // Calling function document.write(calculate(x, k, m)); // This code is contributed // by sravan kumar </script> |
2
Time Complexity: O(k * logx), where k and x represents the value of the given integers.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
An efficient solution is to use Euler’s Totient Function to solve this problem. Since x is a prime number and is always greater than m, that means x and m will always be co-prime. So the fact that will help here is (a^b)%m = (a^(b % et(m)))%m, where et(m) is Euler Totient Function. Consider having a function calculate(x, k, m) that gives the value (x^x^x^x…k times)%m. (x^x^x^x…k times)%m can be written as (a^b)%m = (a^(b % et(m)))%m, where b = calculate(x, k-1, et(m)). A recursive function can be written, with the base cases when k=0 then, answer is 1, and if m=1, then answer is 0.
Below is the implementation of the above approach.
// C++ program to compute // x^x^x^x.. %m #include <bits/stdc++.h> using namespace std;
const int N = 1000000;
// Create an array to store // phi or totient values long long phi[N + 5];
// Function to calculate Euler // Totient values void computeTotient()
{ // indicates not evaluated yet
// and initializes for product
// formula.
for ( int i = 1; i <= N; i++)
phi[i] = i;
// Compute other Phi values
for ( int p = 2; p <= N; p++) {
// If phi[p] is not computed already,
// then number p is prime
if (phi[p] == p) {
// Phi of a prime number p is
// always equal to p-1.
phi[p] = p - 1;
// Update phi values of all
// multiples of p
for ( int i = 2 * p; i <= N; i += p) {
// Add contribution of p to its
// multiple i by multiplying with
// (1 - 1/p)
phi[i] = (phi[i] / p) * (p - 1);
}
}
}
} // Iterative Function to calculate (x^y)%p in O(log y) long long power( long long x, long long y, long long p)
{ long long res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0) {
// If y is odd, multiply x with result
if (y & 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
} // Function to calculate // (x^x^x^x...k times)%m long long calculate( long long x, long long k,
long long mod)
{ // to store different mod values
long long arr[N];
long long count = 0;
while (mod > 1) {
arr[count++] = mod;
mod = phi[mod];
}
long long result = 1;
long long loop = count + 1;
arr[count] = 1;
// run loop in reverse to calculate
// result
for ( int i = min(k, loop) - 1; i >= 0; i--)
result = power(x, result, arr[i]);
return result;
} // Driver Code int main()
{ // compute euler totient function values
computeTotient();
long long x = 3, k = 2, m = 3;
// Calling function to compute answer
cout << calculate(x, k, m) << endl;
return 0;
} |
// C program to compute // x^x^x^x.. %m #include <stdio.h> #define N 1000000 // Create an array to store // phi or totient values long long phi[N + 5];
// Function to calculate Euler // Totient values int min( int a, int b)
{ int min = a;
if (min > b)
min = b;
return min;
} void computeTotient()
{ // indicates not evaluated yet
// and initializes for product
// formula.
for ( int i = 1; i <= N; i++)
phi[i] = i;
// Compute other Phi values
for ( int p = 2; p <= N; p++) {
// If phi[p] is not computed already,
// then number p is prime
if (phi[p] == p) {
// Phi of a prime number p is
// always equal to p-1.
phi[p] = p - 1;
// Update phi values of all
// multiples of p
for ( int i = 2 * p; i <= N; i += p) {
// Add contribution of p to its
// multiple i by multiplying with
// (1 - 1/p)
phi[i] = (phi[i] / p) * (p - 1);
}
}
}
} // Iterative Function to calculate (x^y)%p in O(log y) long long power( long long x, long long y, long long p)
{ long long res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0) {
// If y is odd, multiply x with result
if (y & 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
} // Function to calculate // (x^x^x^x...k times)%m long long calculate( long long x, long long k,
long long mod)
{ // to store different mod values
long long arr[N];
long long count = 0;
while (mod > 1) {
arr[count++] = mod;
mod = phi[mod];
}
long long result = 1;
long long loop = count + 1;
arr[count] = 1;
// run loop in reverse to calculate
// result
for ( int i = min(k, loop) - 1; i >= 0; i--)
result = power(x, result, arr[i]);
return result;
} // Driver Code int main()
{ // compute euler totient function values
computeTotient();
long long x = 3, k = 2, m = 3;
// Calling function to compute answer
printf ( "%lld\n" ,calculate(x, k, m));
return 0;
} // This code is contributed by kothavvsaakash. |
// Java program for computing // x^x^x^x.. %m class GFG
{ // Create an array to store // phi or totient values static int N = 1000000 ;
static long phi[] = new long [N + 5 ];
// Function to calculate // Euler Totient values static void computeTotient()
{ // indicates not evaluated
// yet and initializes for
// product formula.
for ( int i = 1 ; i <= N; i++)
phi[i] = i;
// Compute other Phi values
for ( int p = 2 ; p <= N; p++)
{
// If phi[p] is not
// computed already,
// then number p is prime
if (phi[p] == p)
{
// Phi of a prime number p
// is always equal to p-1.
phi[p] = p - 1 ;
// Update phi values of
// all multiples of p
for ( int i = 2 * p; i <= N; i += p)
{
// Add contribution of p
// to its multiple i by
// multiplying with (1 - 1/p)
phi[i] = (phi[i] / p) *
(p - 1 );
}
}
}
} // Iterative Function to // calculate (x^y)%p in O(log y) static long power( long x, long y, long p)
{ long res = 1 ; // Initialize result
x = x % p; // Update x if it is
// more than or equal to p
while (y > 0 )
{
// If y is odd, multiply
// x with result
if ((y & 1 ) > 0 )
res = (res * x) % p;
// y must be even now
y = y >> 1 ; // y = y/2
x = (x * x) % p;
}
return res;
} // Function to calculate // (x^x^x^x...k times)%m static long calculate( long x, long k,
long mod)
{ // to store different
// mod values
long arr[] = new long [N];
long count = 0 ;
while (mod > 1 )
{
arr[( int )count++] = mod;
mod = phi[( int )mod];
}
long result = 1 ;
long loop = count + 1 ;
arr[( int )count] = 1 ;
// run loop in reverse
// to calculate result
for ( int i = ( int )Math.min(k, loop) - 1 ;
i >= 0 ; i--)
result = power(x, result, arr[i]);
return result;
} // Driver Code public static void main(String args[])
{ // compute euler totient
// function values
computeTotient();
long x = 3 , k = 2 , m = 3 ;
// Calling function
// to compute answer
System.out.println(calculate(x, k, m));
} } // This code is contributed by Arnab Kundu |
# Python3 program to compute # x^x^x^x.. %m N = 1000000
# Create an array to store # phi or totient values phi = [ 0 for i in range (N + 5 )]
# Function to calculate Euler # Totient values def computeTotient():
# indicates not evaluated yet
# and initializes for product
# formula.
for i in range ( 1 , N + 1 ):
phi[i] = i
# Compute other Phi values
for p in range ( 2 , N + 1 ):
# If phi[p] is not computed already,
# then number p is prime
if (phi[p] = = p):
# Phi of a prime number p is
# always equal to p-1.
phi[p] = p - 1
# Update phi values of all
# multiples of p
for i in range ( 2 * p, N + 1 , p):
# Add contribution of p to its
# multiple i by multiplying with
# (1 - 1/p)
phi[i] = (phi[i] / / p) * (p - 1 )
# Iterative Function to calculate (x^y)%p in O(log y) def power(x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more than or
# equal to p
while (y > 0 ):
# If y is odd, multiply x with result
if (y & 1 ):
res = (res * x) % p
# y must be even now
y = y >> 1 # y = y/2
x = (x * x) % p
return res
# Function to calculate # (x^x^x^x...k times)%m def calculate(x, k,mod):
# to store different mod values
arr = [ 0 for i in range (N)]
count = 0
while (mod > 1 ):
arr[count] = mod
count + = 1
mod = phi[mod]
result = 1
loop = count + 1
arr[count] = 1
# run loop in reverse to calculate
# result
for i in range ( min (k,loop), - 1 , - 1 ):
result = power(x, result, arr[i])
return result
# Driver Code # compute euler totient function values computeTotient() x = 3
k = 2
m = 3
# Calling function to compute answer print (calculate(x, k, m))
# This code is contributed by mohit kumar 29 |
// C# program for computing // x^x^x^x.. %m using System;
class GFG
{ // Create an array to store // phi or totient values static int N = 1000000;
static long []phi = new long [N + 5];
// Function to calculate // Euler Totient values static void computeTotient()
{ // indicates not evaluated
// yet and initializes for
// product formula.
for ( int i = 1; i <= N; i++)
phi[i] = i;
// Compute other Phi values
for ( int p = 2; p <= N; p++)
{
// If phi[p] is not
// computed already,
// then number p is prime
if (phi[p] == p)
{
// Phi of a prime
// number p is
// always equal
// to p-1.
phi[p] = p - 1;
// Update phi values
// of all multiples
// of p
for ( int i = 2 * p;
i <= N; i += p)
{
// Add contribution of p
// to its multiple i by
// multiplying with (1 - 1/p)
phi[i] = (phi[i] / p) *
(p - 1);
}
}
}
} // Iterative Function to // calculate (x^y)%p in O(log y) static long power( long x,
long y, long p)
{ long res = 1; // Initialize result
x = x % p; // Update x if it
// is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply
// x with result
if ((y & 1) > 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
} // Function to calculate // (x^x^x^x...k times)%m static long calculate( long x, long k,
long mod)
{ // to store different
// mod values
long []arr = new long [N];
long count = 0;
while (mod > 1)
{
arr[( int )count++] = mod;
mod = phi[( int )mod];
}
long result = 1;
long loop = count + 1;
arr[( int )count] = 1;
// run loop in reverse
// to calculate result
for ( int i = ( int )Math.Min(k, loop) - 1;
i >= 0; i--)
result = power(x, result,
arr[i]);
return result;
} // Driver Code static public void Main ()
{ // compute euler totient // function values computeTotient(); long x = 3, k = 2, m = 3;
// Calling function // to compute answer Console.WriteLine(calculate(x, k, m)); } } // This code is contributed // by akt_mit |
<script> // Javascript program for computing x^x^x^x.. %m
// Create an array to store
// phi or totient values
let N = 1000000;
let phi = new Array(N + 5);
phi.fill(0);
// Function to calculate
// Euler Totient values
function computeTotient()
{
// indicates not evaluated
// yet and initializes for
// product formula.
for (let i = 1; i <= N; i++)
phi[i] = i;
// Compute other Phi values
for (let p = 2; p <= N; p++)
{
// If phi[p] is not
// computed already,
// then number p is prime
if (phi[p] == p)
{
// Phi of a prime number p
// is always equal to p-1.
phi[p] = p - 1;
// Update phi values of
// all multiples of p
for (let i = 2 * p; i <= N; i += p)
{
// Add contribution of p
// to its multiple i by
// multiplying with (1 - 1/p)
phi[i] = (phi[i] / p) * (p - 1);
}
}
}
}
// Iterative Function to
// calculate (x^y)%p in O(log y)
function power(x, y, p)
{
let res = 1; // Initialize result
x = x % p; // Update x if it is
// more than or equal to p
while (y > 0)
{
// If y is odd, multiply
// x with result
if ((y & 1) > 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Function to calculate
// (x^x^x^x...k times)%m
function calculate(x, k, mod)
{
// to store different
// mod values
let arr = new Array(N);
arr.fill(0);
let count = 0;
while (mod > 1)
{
arr[count++] = mod;
mod = phi[mod];
}
let result = 1;
let loop = count + 1;
arr[count] = 1;
// run loop in reverse
// to calculate result
for (let i = Math.min(k, loop) - 1; i >= 0; i--)
result = power(x, result, arr[i]);
return result;
}
// compute euler totient
// function values
computeTotient();
let x = 3, k = 2, m = 3;
// Calling function
// to compute answer
document.write(calculate(x, k, m));
// This code is contributed by rameshtravel07. </script> |
0
Time Complexity: O(N), where N is 106 since all the Euler Totient values are pre-calculated.
Auxiliary Space: O(N), where N is 106