（Problem 35）Circular primes

The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.

There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.

How many circular primes are there below one million?

题目大意：

我们称197为一个循环质数，因为它的所有轮转形式: 197, 971和719都是质数。

100以下有13个这样的质数: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, 和97.

100万以下有多少个循环质数？

//（Problem 35）Circular primes

// Completed on Fri, 26 Jul 2013, 06:17

// 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;

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

    {

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

    }

    return true;

}



bool circular_prime(int n)

{

    int i,j,flag=1;

    char s[6];

    int sum=0;

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

    int len=strlen(s);

    for(i=0; i<len; i++)

    {

        if(s[i]!='1' && s[i]!='3' && s[i]!='7' && s[i]!='9')

            return false;

    }

    for(i=0; i<len; i++)

    {

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

        {

            sum+=s[j%len]-'0';

            sum*=10;

        }

        sum+=s[j%len]-'0';

        if(!isprim(sum)) return false;

        sum=0;

    }

    return true;

}





int main()

{

    int sum=4;    //已包含2,3,5,7

    for(int i=11; i<1000000; i++)

    {

        if(circular_prime(i))   

            sum++;

    }

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

    return 0;

}

Answer:

55