Circular Prime Number

PROBLEM :
A prime number is a Circular Prime Number if all of its possible rotations are itself prime numbers. Now given a number N check if it is Circular Prime or Not.

Input:
The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case contains an integer N.

Output:
For each test case in a new line print 1 if the number is circular prime else 0.

Constraints:
1<=T<=100
1<=N<=104

Example:
Input:
2
197
101

Output:
1
0

Explanation:
For first test case:
197 is a Circular Prime because all rotations of 197 are
197, 719, 971 all of the 3 are prime number's
hence 197 is a circular prime

For second test case:
101 isn't a prime number as
rotation of 101 yields 110 which isn't a prime number
hence 101 isn't circular prime

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SIMPLE c++ IMPLEMENTATION :
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#include<iostream>
using namespace std;
#include<math.h>
#define max 100000

int Circular_Prime_Number(int ,int [],int) ;
void Sieve_of_Eratosthenes(int [],int ) ;
int No_digits(int ) ;
int rotate_no(int ,int ) ;

int main()
 {
    int arr[max] ;
Sieve_of_Eratosthenes(arr,max) ;
int t,no ;
cin>>t ;

while(t--)
{
   cin>>no ;
   cout<<Circular_Prime_Number(no,arr,max)<<endl ;
}
return 0;
}

int Circular_Prime_Number(int no,int arr[],int k)
{
    if(no==0||no==1)
        return 0 ;
       
    int i,j,count,N ;
    count=No_digits(no) ;
    N=count ;
   
    while(count--)
    {
        if(arr[no]==0)
            return 0 ;
       
        no=rotate_no(no,N) ;
        if(no==0)
            return 0 ;
    }
    return 1 ;
}

void Sieve_of_Eratosthenes(int arr[],int no)
{
    int i,j;
    for(i=0;i<=no;i++)
        arr[i]=1 ;
       
    arr[0]=0 ;
    arr[1]=0 ;
   
    for(i=2;i<=sqrt(no);i++)
    {
    if(arr[i]==1)
        for(j=i+i;j<=no;j+=i)
            arr[j]=0 ;
    }
}

int No_digits(int no)
{
    int i,count ;
    count=0 ;
   
    while(no)
    {
        count++ ;
        no=no/10 ;
    }
    return count ;
}

int rotate_no(int no,int N)
{
    int r ;
   
    r=no%10 ;
   
    if(r==0)
        return 0 ;
       
    no=no/10 ;
    no=pow(10,N-1)*r+no ;
    return no ;
}

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