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## Project Euler 35–Circular primes

Wu Yudong    August 20, 2018     欧拉计划   420

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?

197被称为圆周素数，因为将它逐位旋转所得到的数：197，971和719都是素数。

//（Problem 35）Circular primes
// Completed on Fri, 26 Jul 2013, 06:17
// Language: C
//
// 版权所有（C）wu yudong
// 博客地址：http://www.wuyudong.com
#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;
}