Comprobar si un número es Euler Pseudoprime

Dado un entero N y un número de base A , la tarea es verificar si N es un pseudoprimo de Euler para la base A dada . 
Un entero N se llama Euler Pseudoprim a la base A , si 
 

  1. A > 0 y N es un número compuesto impar.
  2. A y N son coprimos, es decir , GCD(A, N) = 1 .
  3. A (N – 1) / 2 % N es 1 o N – 1 .

Ejemplos: 
 

Entrada: N = 121, A = 3 
Salida:
Entrada: N = 343, A = 2 
Salida: No 
 

Enfoque: compruebe todas las condiciones dadas para Euler Pseudoprime. Si alguna de las condiciones no es verdadera, imprima No; de lo contrario, imprima Sí.
A continuación se muestra la implementación del enfoque anterior: 
 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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

Javascript

<script>
 
// JavaScript implementation of the approach
 
// Function that returns true if n is composite
function 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 (let 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)
function Power(x, y, p) {
 
    // Initialize result
    let 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 __gcd(a, b) {
    if (b == 0)
        return a;
    return __gcd(b, a % b);
}
 
// Function that returns true if N is
// Euler Pseudoprime to the base A
function isEulerPseudoprime(N, 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;
 
    let 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
 
let N = 121, A = 3;
 
if (isEulerPseudoprime(N, A))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by _saurabh_jaiswal
 
</script>
Producción: 

Yes

 

Complejidad del tiempo: O(sqrt(N))

Espacio Auxiliar: O(1)
 

Publicación traducida automáticamente

Artículo escrito por harshit23 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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