（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: acutus@126.com) 
// 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