计算功率k乘以％m

给定x，k和m。计算(x ^xxx…k )％m，x的幂为k倍。给定x始终为质数，m大于x。
例子：

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

天真的方法是计算xk次的幂并每次都执行模运算。

C++

// C++ program for computing
// x^x^x^x.. %m
#include 
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;
}

Java

// 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

# 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#

// 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

 $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
?>

Javascript

//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

输出：

2

一个有效的解决方案是使用欧拉的Totient函数来解决此问题。由于x是质数，并且始终大于m，这意味着x和m将始终是互质数。因此，这里有帮助的事实是(a ^ b)％m =(a ^(b％et(m)))％m ，其中et(m)是Euler Totient 函数。考虑具有一个函数calculate(x，k，m) ，该函数给出值(x ^ x ^ x ^ x…k倍)％m 。 (x ^ x ^ x ^ x…k次)％m可以写成(a ^ b)％m =(a ^(b％et(m)))％m ，其中b =计算(x，k- 1，et(m)) 。可以编写一个递归函数，其基本情况是k = 0时，答案为1；如果m = 1，则答案为0。
下面是上述方法的实现。

C++

// C++ program to compute
// x^x^x^x.. %m
#include 
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;
}

Java

// 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

# 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＃

// 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

输出：

0

时间复杂度：O(N)，因为所有的Euler Totent值都已预先计算，所以N为10 ⁶ 。
辅助空间： O(N)，其中N是10 ⁶