Project Euler 問題37（切り捨て可能素数）を解いてみる（続き）

#include <stdio.h>

#define N 100000000
unsigned char is_prime_table[N+1];
#define is_prime(n) (is_prime_table[(n)])

void generate_is_prime_table(int n)
{
	int i, j;

	for (i = 2; i <= n; ++i) {
		is_prime_table[i] = 1;
	}

	for (i = 2; i * i <= n; ++i) {
		if (is_prime_table[i] == 1) {
			for (j = i + i; j <= n; j += i) {
				is_prime_table[j] = 0;
			}
		}
	}

	return;
}

int is_satisfied(int number)
{
	int ret_val = 1;
	int original, digits;
	int i, mask, check_number;

	original = number;

	digits = 0;
	while (number > 0) {
		++digits;
		number = number / 10;
	}

	mask = 1;
	for (i = 2; i <= digits; ++i) {
		mask = mask * 10;
	}

	if (is_prime(original) == 0) {
		ret_val = 0;
	}

	check_number = original % mask;
	while (mask >= 10) {
		if (is_prime(check_number) == 0) {
			ret_val = 0;
			break;
		}
		mask = mask / 10;
		check_number = original % mask;
	}

	check_number = original / 10;
	while (check_number > 0) {
		if (is_prime(check_number) == 0) {
			ret_val = 0;
			break;
		}
		check_number = check_number / 10;
	}

	return ret_val;
}

int main(void)
{
	int n;

	generate_is_prime_table(N);

	for (n = 10; n < N; ++n) {
		if (is_satisfied(n) == 1) {
			printf("%d\n", n);
		}
	}

	return 0;
}