﻿ C++ : Find circular prime numbers upto a specific limit

# C++ Exercises: Find circular prime numbers upto a specific limit

## C++ Numbers: Exercise-31 with Solution

Write a program in C++ to find circular prime numbers upto a specific limit.

Sample Solution:

C++ Code :

``````#include<iostream>
#include<cmath>

using namespace std;

int flg;
void chkPrime(long int n)
{
long int i;
i=n-1;
while(i>=2)
{
if(n%i==0)
{
flg=1;
}
i--;
}
}

void AllCombination(long int a)
{
long int b,c,d,e,i,j,k,s,z,v,x[8],y[8],m;
b=a;
i=0;
while(b>0)
{
y[i]=b % 10;
b=b/10;
i++;
}
c=0;
for(j=i-1;j>=0;j--)
{
x[c]=y[j];
c++;
}
m=i;
while(m>0)
{
c=m-1;
d=i-1;
e=0;
s=0;
while(e<i)
{
z=pow(10,d);
v=z*x[c%i];
c++;
d--;
e++;
s=s+v;
}
m--;
chkPrime(s);
}
}

int main()
{
long int i=2,ctr;

cout << "\n\n Find Circular Prime Numbers upto a specific limit: \n";
cout << " ---------------------------------------------------\n";
cout<<" Enter the upper Limit: ";
cin>>ctr;
cout<<"\n The Circular Prime Numbers less than "<<ctr<<" are: "<<endl;
while(i<=ctr)
{
flg=0;
AllCombination(i);
if(flg==0)
{
cout<<i<<" ";
}
i++;
}
cout<<endl;

}
``````

OR

C++ Code:

``````
#include <bits/stdc++.h>
using namespace std;

void SieveOfSundaram (bool marked[], int);
int Rotate(int );
int countDigits(int );

void circularPrime(int n)
{
int nNew = (n-2)/2;
bool marked[nNew + 1];
memset(marked, false, sizeof(marked));
SieveOfSundaram(marked, nNew);
cout << "2 " ;
for (int i=1; i<=nNew; i++)
{
if (marked[i] == true)
continue;
int num = 2 * i + 1;
num = Rotate(num); // function for rotation of prime
while (num != 2*i + 1)
{
if (num % 2 == 0) // check for even
break;
if (marked[(num - 1)/2] == false)
num = Rotate(num);
else
break;
}
if (num == (2*i + 1))
cout << num << " ";
}
}

void SieveOfSundaram(bool marked[], int nNew)
{
for (int i=1; i<=nNew; i++)
for (int j=i; (i + j + 2*i*j) <= nNew; j++)
marked[i + j + 2*i*j] = true;
}
int Rotate(int n)
{
int rem = n % 10;          // find unit place number
rem *= pow(10, countDigits(n)); // to put unit place
n /= 10;                   // remove unit digit
n += rem;                  // add first digit to rest
return n;
}

int countDigits(int n)
{
int digit = 0;
while (n /= 10)
digit++;
return digit;
}

int main(void)
{
int n = 100;
circularPrime(n);
return 0;
}
``````

Sample Output:

```Find Circular Prime Numbers upto a specific limit:
---------------------------------------------------
Enter the upper Limit: 248

The Circular Prime Numbers less than 248 are:
2 3 5 7 11 13 17 31 37 71 73 79 97 113 131 197 199
```

Flowchart:

C++ Code Editor:

What is the difficulty level of this exercise?

﻿

## C++ Programming: Tips of the Day

How to use the PI constant in C++?

On some (especially older) platforms (see the comments below) you might need to

#define _USE_MATH_DEFINES and then include the necessary header file:
`#include<math.h>`

and the value of pi can be accessed via:

`M_PI`

In math.h (2014) it is defined as:

`# define M_PI           3.14159265358979323846  /* pi */`

but check your math.h for more. An extract from the "old" math.h (in 2009):

```/* Define _USE_MATH_DEFINES before including math.h to expose these macro
* definitions for common math constants.  These are placed under an #ifdef
* since these commonly-defined names are not part of the C/C++ standards.
*/```

However:

1. on newer platforms (at least on my 64 bit Ubuntu 14.04) I do not need to define the _USE_MATH_DEFINES
2. On (recent) Linux platforms there are long double values too provided as a GNU Extension:
`# define M_PIl          3.14159265358979323846`

Ref: https://bit.ly/3G4BgzQ