sicily 4190 Prime Palindromes

题目链接：http://soj.me/4190

要点：1 判断回文不能用字符串 2 判断质数时外循环终止条件用sqrt(MAX)  3判断质数时内循环的起始条件用i * i  4将回文质数全部存储出来

#include<iostream>
#include<string>
#include<cmath>
#include<memory.h>
const int MAX = 10000005;
bool visited[MAX];
int result[MAX];
int num = 0;
using namespace std;
bool judge(int a)
{
     int t = a;
     int b = 0;
     while(a != 0)
     {
             b = b * 10 + a % 10; 
             a = a / 10;
     }
     return t == b;
}

void judge2()
{
     for(int i = 2 ; i <= sqrt(MAX) ; i ++)
             if(!visited[i])
             {
                            for(int j = i * i ; j <= MAX ; j += i)
                                    visited[j] = true;
             }                     
     for(int i = 2 ; i < MAX ; i ++)
             if(!visited[i] && judge(i))
                            result[num ++] = i;
}

int main()
{
    int a,b;
    memset(visited,0,sizeof(visited));
    judge2();
    cin >> a >> b;
    while(a != 0 && b != 0)
    {
            for(int i = 0 ; i < num ; i ++)
                    if(result[i] >= a && result[i] <= b)
                                   cout << result[i] << endl;
            cin >> a >> b;
    }
}