Prime Palindromes

　　博客园速度非常不稳定，可能要考虑换地方了。虽然我非常喜欢博客园的模板和气氛。

　　这个题早就知道是怎么做的了。先求出回文数在再判断是不是素数。关键是不知道区间，那就把所有的全部求出来。虽然可能会超时，但是如果使用点技巧的话还是没问题的。

　　如果先筛素数的话开一亿的数组90m，肯定超内存了。据说可以只开一千万的，因为偶数就不用判了嘛。这个具体怎么实现尚不知晓。usaco的判题机构貌似极其严格，就连伪素数测试都能挑出错来。就直接拿模板上的素性测试过了。自己写的暴力测素数函数超了。代码贴一下。日后再整理吧。这两天都没怎么写解题报告了。

/*
ID: like_091
PROG: pprime
LANG: C++
*/
#include<iostream>
#include<fstream>
#include<string>
#include<cmath>
using namespace std;
const int MAX = 15000;
int p[MAX], k = 0;
bool is_prime(int u)
{
	if (u == 0 || u == 1) return false;
	if (u == 2) return true;
	if (u % 2 == 0) return false;
	for (int i = 3; i <= sqrt(u); i += 2)
	if (u % i ==0) return false;
	return true;
}
void add()
{
	int d[4] = {1, 3, 7, 9};
	p[k++] = 5;
	p[k++] = 7;
	p[k++] = 11;
	for (short  i = 0; i < 4; i++)
		for (short  j = 0; j < 10; j++)
			p[k++] = d[i] * 101 + j * 10;
	for (short i = 0; i < 4; i++)
		for (short j = 0; j < 10; j++)
			p[k++] = d[i] * 1001 + j * 110;
	for (short i = 0; i < 4; i++)
		for (short j = 0; j < 10; j++)
			for (short l = 0; l < 10; l++)
				p[k++] = d[i] * 10001 + j * 1010 + l * 100;
	for (short i = 0; i < 4; i++)
		for (short j = 0; j < 10; j++)
			for (short l = 0; l < 10; l++)
				p[k++] = d[i] * 100001 + j * 10010 + l * 1100;
	for (short i = 0; i < 4; i++)
		for (short j = 0; j < 10; j++)
			for (short l = 0; l < 10; l++)
				for (short m = 0; m < 10; m++)
					p[k++] = d[i] * 1000001 + j * 100010 + l * 10100 + m * 1000;
	for (short i = 0; i < 4; i++)
		for (short j = 0; j < 10; j++)
			for (short l = 0; l < 10; l++)
				for (short m = 0; m < 9; m++)
					p[k++] = d[i] * 10000001 + j * 1000010 + l * 100100 + 11000;
}
int main()
{
	ifstream cin("pprime.in");
	ofstream cout("pprime.out");
	add();
	int a, b;
	cin>>a>>b;
	int q = 0;
	while (p[q] < a)++q;
	while (q < k)
	{
		if (p[q] <= b && is_prime(p[q]))
			cout<<p[q]<<endl;
		q++;
	}
	return 0;
}