检查数字是否为Euler Pseudoprime

给定一个整数N和一个基数A ，任务是检查N是否是给定基数A的欧拉伪素数。
整数N被称为以底数A为底的Euler伪素数

1. A> 0 ， N是一个奇数的复合数。
2. A和N是互质的，即GCD(A，N)= 1 。
3. ^{(N – 1)/ 2} ％N为1或N – 1 。

例子：

Input: N = 121, A = 3
Output: Yes

Input: N = 343, A = 2
Output: No

方法：检查所有给定的欧拉伪素条件。如果任一条件不成立，则打印“否”，否则打印“是”。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function that returns true if n is composite
bool isComposite(int n)
{
    // Check if there is any divisor of n.
    // we only need check divisor till sqrt(n)
    // because if there is divisor which is greater
    // than sqrt(n) then there must be a divisor
    // which is less than sqrt(n)
    for (int i = 2; i <= sqrt(n); i++) {
        if (n % i == 0)
            return true;
    }
    return false;
}
  
// Iterative Function to calculate
// (x^y) % p in O(log y)
int Power(int x, int y, int p)
{
  
    // Initialize result
    int res = 1;
  
    // Update x if it is greater
    // than or equal to p
    x = x % 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 that returns true if N is
// Euler Pseudoprime to the base A
bool isEulerPseudoprime(int N, int A)
{
  
    // Invalid base
    if (A <= 0)
        return false;
  
    // N is not a composite odd number
    if (N % 2 == 0 || !isComposite(N))
        return false;
  
    // If A and N are not coprime
    if (__gcd(A, N) != 1)
        return false;
  
    int mod = Power(A, (N - 1) / 2, N);
    if (mod != 1 && mod != N - 1)
        return false;
  
    // All the conditions for Euler
    // Pseudoprime are satisfied
    return true;
}
  
// Driver code
int main()
{
  
    int N = 121, A = 3;
  
    if (isEulerPseudoprime(N, A))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

Java

// Java implementation of the approach
class GFG 
{
  
// Function that returns true if n is composite
static boolean isComposite(int n)
{
    // Check if there is any divisor of n.
    // we only need check divisor till sqrt(n)
    // because if there is divisor which is greater
    // than sqrt(n) then there must be a divisor
    // which is less than sqrt(n)
    for (int i = 2; i <= Math.sqrt(n); i++) 
    {
        if (n % i == 0)
            return true;
    }
    return false;
}
  
// Iterative Function to calculate
// (x^y) % p in O(log y)
static int Power(int x, int y, int p)
{
  
    // Initialize result
    int res = 1;
  
    // Update x if it is greater
    // than or equal to p
    x = x % p;
  
    while (y > 0)
    {
  
        // If y is odd, multiply x with result
        if (y % 2 == 1)
        {
            res = (res * x) % p;
        }
  
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
  
// Function that returns true if N is
// Euler Pseudoprime to the base A
static boolean isEulerPseudoprime(int N, int A)
{
  
    // Invalid base
    if (A <= 0)
        return false;
  
    // N is not a composite odd number
    if (N % 2 == 0 || !isComposite(N))
        return false;
  
    // If A and N are not coprime
    if (__gcd(A, N) != 1)
        return false;
  
    int mod = Power(A, (N - 1) / 2, N);
    if (mod != 1 && mod != N - 1)
        return false;
  
    // All the conditions for Euler
    // Pseudoprime are satisfied
    return true;
}
  
static int __gcd(int a, int b) 
{ 
    if (b == 0) 
        return a; 
    return __gcd(b, a % b); 
      
} 
  
// Driver code
public static void main(String []args) 
{
    int N = 121, A = 3;
  
    if (isEulerPseudoprime(N, A))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 program for nth Fuss–Catalan Number 
from math import gcd, sqrt
  
# Function that returns true if n is composite 
def isComposite(n) : 
  
    # Check if there is any divisor of n. 
    # we only need check divisor till sqrt(n) 
    # because if there is divisor which is greater 
    # than sqrt(n) then there must be a divisor 
    # which is less than sqrt(n) 
    for i in range(2, int(sqrt(n)) + 1) : 
        if (n % i == 0) :
            return True; 
  
    return False; 
  
# Iterative Function to calculate 
# (x^y) % p in O(log y) 
def Power(x, y, p) : 
  
    # Initialize result 
    res = 1; 
  
    # Update x if it is greater 
    # than or equal to p 
    x = x % 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 that returns true if N is 
# Euler Pseudoprime to the base A 
def isEulerPseudoprime(N, A) :
  
    # Invalid base 
    if (A <= 0) :
        return False; 
  
    # N is not a composite odd number 
    if (N % 2 == 0 or not isComposite(N)) :
        return False; 
  
    # If A and N are not coprime 
    if (gcd(A, N) != 1) :
        return false; 
  
    mod = Power(A, (N - 1) // 2, N); 
    if (mod != 1 and mod != N - 1) :
        return False; 
  
    # All the conditions for Euler 
    # Pseudoprime are satisfied 
    return True; 
  
# Driver code 
if __name__ == "__main__" : 
  
    N = 121; A = 3; 
  
    if (isEulerPseudoprime(N, A)) :
        print("Yes"); 
    else :
        print("No");
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System; 
  
class GFG
{
  
// Function that returns true if n is composite
static bool isComposite(int n)
{
    // Check if there is any divisor of n.
    // we only need check divisor till sqrt(n)
    // because if there is divisor which is greater
    // than sqrt(n) then there must be a divisor
    // which is less than sqrt(n)
    for (int i = 2; i <= Math.Sqrt(n); i++) 
    {
        if (n % i == 0)
            return true;
    }
    return false;
}
  
// Iterative Function to calculate
// (x^y) % p in O(log y)
static int Power(int x, int y, int p)
{
  
    // Initialize result
    int res = 1;
  
    // Update x if it is greater
    // than or equal to p
    x = x % p;
  
    while (y > 0)
    {
  
        // If y is odd, multiply x with result
        if (y % 2 == 1)
        {
            res = (res * x) % p;
        }
  
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
  
// Function that returns true if N is
// Euler Pseudoprime to the base A
static bool isEulerPseudoprime(int N, int A)
{
  
    // Invalid base
    if (A <= 0)
        return false;
  
    // N is not a composite odd number
    if (N % 2 == 0 || !isComposite(N))
        return false;
  
    // If A and N are not coprime
    if (__gcd(A, N) != 1)
        return false;
  
    int mod = Power(A, (N - 1) / 2, N);
    if (mod != 1 && mod != N - 1)
        return false;
  
    // All the conditions for Euler
    // Pseudoprime are satisfied
    return true;
}
  
static int __gcd(int a, int b) 
{ 
    if (b == 0) 
        return a; 
    return __gcd(b, a % b); 
      
} 
  
// Driver code
public static void Main(String []args) 
{
    int N = 121, A = 3;
  
    if (isEulerPseudoprime(N, A))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
  
// This code is contributed by PrinciRaj1992

输出：

Yes

时间复杂度： O(sqrt(N))

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