Compute power of power k times % m

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

2

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++

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// 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;
}

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Java

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// 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

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C#

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// 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

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Output:

0

Time Complexity: O(N), where N is 106 since all the Euler Totent values are pre-calculated.



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