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++
// C++ program for computing// x^x^x^x.. %m#include <bits/stdc++.h>using namespace std;// Function to compute the given valueint 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 Codeint main(){ int x = 5, k = 2, m = 3; // Calling function cout << calculate(x, k, m); return 0;} |
C
// C program for computing// x^x^x^x.. %m#include <stdio.h>#include <math.h>// Function to compute the given valueint 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 Codeint 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
// Java program for computing// x^x^x^x.. %mclass GFG{// Function to compute// the given valuestatic 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 Codepublic 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
# Python3 program for # computing x^x^x^x.. %mimport math# Function to compute# the given valuedef 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 Codex = 5;k = 2;m = 3;# Calling functionprint(calculate(x, k, m)); # This code is contributed # by mits |
C#
// C# program for computing// x^x^x^x.. %musing System;class GFG{ // Function to compute// the given valuestatic 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 Codestatic 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// PHP program for computing// x^x^x^x.. %m// Function to compute// the given valuefunction 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 functionecho calculate($x, $k, $m); // This code is contributed // by akt_mit?> |
Javascript
<script>//program for computing// x^x^x^x.. %m// Function to compute// the given valuefunction 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 Codelet x = 5;let k = 2;let m = 3;// Calling functiondocument.write(calculate(x, k, m)); // This code is contributed// by sravan kumar</script> |
Output:
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++
// 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 valueslong long phi[N + 5];// Function to calculate Euler// Totient valuesvoid 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)%mlong 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 Codeint 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
// C program to compute// x^x^x^x.. %m#include <stdio.h>#define N 1000000// Create an array to store// phi or totient valueslong long phi[N + 5];// Function to calculate Euler// Totient valuesint 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)%mlong 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 Codeint 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
// Java program for computing// x^x^x^x.. %mclass GFG{// Create an array to store// phi or totient valuesstatic int N = 1000000;static long phi[] = new long[N + 5];// Function to calculate // Euler Totient valuesstatic 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)%mstatic 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 Codepublic 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
# Python3 program to compute# x^x^x^x.. %mN = 1000000# Create an array to store# phi or totient valuesphi=[0 for i in range(N + 5)]# Function to calculate Euler# Totient valuesdef 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)%mdef 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 valuescomputeTotient()x = 3k = 2m = 3# Calling function to compute answerprint(calculate(x, k, m))# This code is contributed by mohit kumar 29 |
C#
// C# program for computing// x^x^x^x.. %musing System;class GFG{ // Create an array to store// phi or totient valuesstatic int N = 1000000;static long []phi = new long[N + 5];// Function to calculate // Euler Totient valuesstatic 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)%mstatic 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 Codestatic public void Main (){ // compute euler totient// function valuescomputeTotient();long x = 3, k = 2, m = 3;// Calling function // to compute answerConsole.WriteLine(calculate(x, k, m));}}// This code is contributed // by akt_mit |
Javascript
<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> |
Output:
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
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