（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.

题目大意：

3797这个数很有趣。它本身是质数，而且如果我们从左边不断地裁去数字，得到的仍然都是质数：3797,797,97,7。而且我们还可以从右向左裁剪：3797,379,37,3，得到的仍然都是质数。

找出全部11个这样从左向右和从右向左都可以裁剪的质数。
注意：2,3,5和7不被认为是可裁剪的质数。

//（Problem 37）Truncatable primes

// Completed on Thu, 31 Oct 2013, 13:12

// Language: C

//

// 版权所有（C）acutus   (mail: [email protected]) 

// 博客地址：http://www.cnblogs.com/acutus/

#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