（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.

 1 #include<stdio.h>

 2 #include<math.h>

 3 #include<string.h>

 4 #include<ctype.h>

 5 #include<stdlib.h>

 6 #include<stdbool.h>

 7 

 8 bool isprim(int n)

 9 {

10     int i=2;

11     if(n==1) return false;

12     for(; i*i<=n; i++)

13     {

14         if(n%i==0)  return false;

15     }

16     return true;

17 }

18 

19 bool truncatable_prime(int n)

20 {

21     int i,j,t,flag=1;

22     char s[6];

23     int sum=0;

24     sprintf(s,"%d",n);

25     int len=strlen(s);

26 

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

28 

29     for(i=1; i<len-1; i++)

30     {

31         t=s[i]-'0';

32         if(t==0 || t==2 || t==4 || t==6 || t==5 || t==8)  return false;

33     }

34     

35     for(i=1; i<len-1; i++)

36     {

37         for(j=i; j<len-1; j++)

38         {

39             sum+=s[j]-'0';

40             sum*=10;

41         }

42         sum+=s[j]-'0';

43         if(!isprim(sum))  return false;

44         sum=0;

45     }

46     j=len-1;

47     i=0;

48     while(j>i)

49     {

50         for(i=0; i<j; i++)

51         {

52             sum+=s[i]-'0';

53             sum*=10;

54         }

55         sum+=s[i]-'0';

56         if(!isprim(sum)) return false;

57         sum=0;

58         i=0;

59         j--;

60     }

61     return true;

62 }

63 

64 int main()

65 {

66     int sum,count;

67     sum=count=0;

68     int i=13;

69     while(1)

70     {

71         if(isprim(i) && truncatable_prime(i))

72         {

73             count++;

74             sum+=i;

75             //printf("%d\n",i);

76         }

77         i=i+2;

78         if(count==11)  break;

79     }

80     printf("%d\n",sum);

81     return 0;

82 }

 

Answer:

748317