https://www.luogu.com.cn/problem/P1217#submit

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>

using  ll = long long int;
using namespace std;

int main()
{

    ll a , b ;
    cin >> a >> b ;

    for(ll i = a ; i <= b ; i++) 
    //第三处优化回文数比质数少 先判断是不是回文 再判断质数
    {

            int cnt = 0;
            ll rever = 0;
            ll temp = i;
            while(temp!= 0)
            {
                cnt = temp%10;
                rever = rever*10 + cnt;
                temp /=10;

            }
            if(rever == i) 
            {
                bool is_prime = true;
                for(ll j = 2 ; j <= i/j ; j++)//第一处优化 判断质数时
                {
                    if(i%j == 0) 
                    {
                        is_prime = false;
                        break; //第二处优化 不是质数直接退循环
                    }
                }
                if(is_prime)
                cout << i << endl;

            }

    }

    return 0;
}