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
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!