PROJECT EULER SOLUTION # 60

Solution to problem number 60 of Project Euler.

Question # 60
The primes 3, 7, 109, and 673, are quite remarkable. By taking any two primes and concatenating them in any order the result will always be prime. For example, taking 7 and 109, both 7109 and 1097 are prime. The sum of these four primes, 792, represents the lowest sum for a set of four primes with this property.
Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime.

Solution # 60

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#include<stdio.h>

#include<math.h>

#include<stdlib.h>

#define max 1000000

int arr[max]={0};

void set_arr();

int is_prime(long int);

int is_conPrime(long int, long int);

int main()

{

      long int i,sum=0,a,b,c,d,e,min=2147483647;

      set_arr();

      for(a=3;a<10000;a+=2)

            if(arr[a])

                  for(b=3;b<10000;b+=2)

                        if(arr[b] && is_conPrime(a,b))

                              for(c=3;c<10000;c+=2)

                                    if(arr[c] && is_conPrime(a,c) && is_conPrime(b,c))

                                          for(d=3;d<10000;d+=2)

                                                if(arr[d] && is_conPrime(a,d) && is_conPrime(b,d) && is_conPrime(c,d))

                                                      for(e=3;e<10000;e+=2)

                                                            if(arr[e] && is_conPrime(a,e) && is_conPrime(b,e) && is_conPrime(c,e) && is_conPrime(d,e))

                                                            {

                                                                  printf("%ld %ld %ld %ld %ld\nsum = %ld",a,b,c,d,e,(a+b+c+d+e));

                                                                  exit(1);

                                                            }

      return 0;

}

void set_arr()

{

      int i;

      for(i=2;i<max;i++)

            if(is_prime(i))

                  arr[i]=1;

}

int is_prime(long int n)

{

      long int i;

      if(n==1)

            return 0;

      if(n==2)

            return 1;

      if(n%2==0)

            return 0;

      for(i=3;i*i<=n;i+=2)

            if(n%i==0)

                  return 0;

      return 1;

}

int count_digits(long int n)

{

      int counter=0;

      while(n)

      {

            counter++;

            n/=10;

      }

      return counter;

}

int is_conPrime(long int a,long int b)

{

      long int num1,num2;

      num1=a+b*(long)pow(10,count_digits(a));

      num2=b+a*(long)pow(10,count_digits(b));

      if(is_prime(num1) && is_prime(num2))

            return 1;

      return 0;

}

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