检查数字是否为素数

我们得到一个数字n。我们的任务是检查数字是否为素数。
循环质数：如果在数字的任何循环排列之后仍保持质数，则质数被称为循环质数。
例子：

Input :  n = 113 
Output : Yes
All cyclic permutations of 113 (311
and 131) are prime. 

Input : 1193
Output : Yes

这个想法很简单，我们生成给定数字的所有排列并检查每个排列的质数。为了生成排列，我们将最后一位逐个移动到第一个位置。

C++

// Program to check if a number is circular
// prime or not.
#include 
#include 
using namespace std;
 
// Function to check if a number is prime or not.
bool isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// Function to check if the number is circular
// prime or not.
bool checkCircular(int N)
{
    // Count digits.
    int count = 0, temp = N;
    while (temp) {
        count++;
        temp /= 10;
    }
 
    int num = N;
    while (isPrime(num)) {
 
        // Following three lines generate the next
        // circular permutation of a number. We move
        // last digit to first position.
        int rem = num % 10;
        int div = num / 10;
        num = (pow(10, count - 1)) * rem + div;
 
        // If all the permutations are checked and
        // we obtain original number exit from loop.
        if (num == N)
            return true;
    }
 
    return false;
}
 
// Driver Program
int main()
{
    int N = 1193;
    if (checkCircular(N))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}

Java

// Java Program to check if a number
// is circular prime or not.
import java.lang.*;
 
class GFG
{
    // Function to check if a number is prime or not.
    static boolean isPrime(int n)
    {
        // Corner cases
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
 
        // This is checked so that we can skip
        // middle five numbers in below loop
        if (n % 2 == 0 || n % 3 == 0)
            return false;
 
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
 
        return true;
    }
 
    // Function to check if the number is circular
    // prime or not.
    static boolean checkCircular(int N)
    {
        // Count digits.
        int count = 0, temp = N;
        while (temp>0) {
            count++;
            temp /= 10;
        }
 
        int num = N;
        while (isPrime(num)) {
 
        // Following three lines generate the next
        // circular permutation of a number. We
        // move last digit to first position.
        int rem = num % 10;
        int div = num / 10;
        num = (int)((Math.pow(10, count - 1)) * rem)
                                             + div;
 
        // If all the permutations are checked and
        // we obtain original number exit from loop.
        if (num == N)
            return true;
        }
 
        return false;
    }
 
    // Driver Program
    public static void main (String[] args)
    {
        int N = 1193;
        if (checkCircular(N))
        System.out.println("Yes");
        else
        System.out.println("No");
    }
}
/* This code is contributed by Kriti Shukla */

Python

# Python Program to check if a number
# is circular prime or not.
 
import math
 
# Function to check if a number is prime
# or not.
def isPrime(n) :
 
    # Corner cases
    if (n <= 1) :
        return False
    if (n <= 3) :
        return True
         
    # This is checked so that we can skip
    # middle five numbers in below loop
    if (n % 2 == 0 or n % 3 == 0) :
        return False
 
    i = 5
    while i * i <= n :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
     
    return True
     
# Function to check if the number is
# circular prime or not.
def checkCircular(N) :
     
    #Count digits.
    count = 0
    temp = N
    while (temp > 0) :
        count = count + 1
        temp = temp / 10
         
    num = N;
    while (isPrime(num)) :
         
        # Following three lines generate the
        # next circular permutation of a
        # number. We move last digit to
        # first position.
        rem = num % 10
        div = num / 10
        num = (int)((math.pow(10, count - 1))
                                * rem)+ div
 
        # If all the permutations are checked
        # and we obtain original number exit
        # from loop.
        if (num == N) :
            return True
     
    return False
     
# Driver Program
N = 1193;
if (checkCircular(N)) :
    print "Yes"
else :
    print "No"
     
# This code is contributed by Nikita Tiwari

C#

// C# Program to check if a number
// is circular prime or not.
using System;
 
class GFG {
     
    // Function to check if a
    // number is prime or not.
    static bool isPrime(int n)
    {
         
        // Corner cases
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
 
        // This is checked so that we
        // can skip middle five numbers
        // in below loop
        if (n % 2 == 0 || n % 3 == 0)
            return false;
 
        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
 
        return true;
    }
 
    // Function to check if the number
    // is circular prime or not.
    static bool checkCircular(int N)
    {
         
        // Count digits.
        int count = 0, temp = N;
        while (temp > 0)
        {
            count++;
            temp /= 10;
        }
 
        int num = N;
        while (isPrime(num))
        {
 
            // Following three lines generate
            // the next circular permutation
            // of a number. We move last digit
            // to first position.
            int rem = num % 10;
            int div = num / 10;
            num = (int)((Math.Pow(10, count -
                         1)) * rem) + div;
     
            // If all the permutations are
            // checked and we obtain original
            // number exit from loop.
            if (num == N)
                return true;
        }
 
        return false;
    }
 
    // Driver code
    public static void Main ()
    {
        int N = 1193;
        if (checkCircular(N))
        Console.Write("Yes");
        else
        Console.Write("No");
    }
}
 
// This code is contributed by Nitin Mittal.

输出：

Yes

时间复杂度： O(N ^1/2 )