（Problem 37）Truncatable primes

The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.

Find the sum of the only eleven primes that are both truncatable from left to right and right to left.

NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<ctype.h>
#include<stdlib.h>
#include<stdbool.h>

bool isprim(int n)
{
    int i=2;
    if(n==1) return false;
    for(; i*i<=n; i++)
    {
        if(n%i==0)  return false;
    }
    return true;
}

bool truncatable_prime(int n)
{
    int i,j,t,flag=1;
    char s[6];
    int sum=0;
    sprintf(s,"%d",n);
    int len=strlen(s);

    if(!isprim(s[0]-'0') || !isprim(s[len-1]-'0')) return false;

    for(i=1; i<len-1; i++)
    {
        t=s[i]-'0';
        if(t==0 || t==2 || t==4 || t==6 || t==5 || t==8)  return false;
    }
    
    for(i=1; i<len-1; i++)
    {
        for(j=i; j<len-1; j++)
        {
            sum+=s[j]-'0';
            sum*=10;
        }
        sum+=s[j]-'0';
        if(!isprim(sum))  return false;
        sum=0;
    }
    j=len-1;
    i=0;
    while(j>i)
    {
        for(i=0; i<j; i++)
        {
            sum+=s[i]-'0';
            sum*=10;
        }
        sum+=s[i]-'0';
        if(!isprim(sum)) return false;
        sum=0;
        i=0;
        j--;
    }
    return true;
}

int main()
{
    int sum,count;
    sum=count=0;
    int i=13;
    while(1)
    {
        if(isprim(i) && truncatable_prime(i))
        {
            count++;
            sum+=i;
            //printf("%d\n",i);
        }
        i=i+2;
        if(count==11)  break;
    }
    printf("%d\n",sum);
    return 0;
}

 

Answer:

748317