题解

因为

11/11=1
1001/11=91
100001/11=9091
10000001/11=909091

所以，$\overline{aa}|11,\overline{abba}|11,\overline{abccba}|11,\overline{abcddcba}|11$，即偶数位的回文质数只有11.
又因为
$5\le a<b\le100,000,000$
所以不存在9位的回文质数.
说完了，开始暴力！

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
vector <int> v;
void _init(){
	v.push_back(11);//提前添加11
	for(int i = 1; i <= 9; i += 2){
		int tmp1 = i;
		v.push_back(tmp1);
		for(int j = 0; j <= 9; j++){
			int tmp2 = i * 100 + j * 10 + i;
			v.push_back(tmp2);
			for(int k = 0; k <= 9; k++){
				int tmp3 = i * 10000 + j * 1000 + k * 100 + j * 10 + i;
				v.push_back(tmp3);
				for(int l = 0; l <= 9; l++){
					int tmp4 = i * 1000000 + j * 100000 + k * 10000 + l * 1000 + k * 100 + j * 10 + i;
					v.push_back(tmp4);//添加回文数
				}
			}
		}
	}sort(v.begin(), v.end());//排序
} 
bool check(int n){//判断
	if(n == 2) return 1;
	if(n < 2 || n % 2 == 0) return 0;
	for(int i = 3; i * i <= n; i += 2){
		if(n % i == 0) return 0;
	}return 1;
}
int main(){
	_init();
	int n, m;
	cin >> n >> m;
	for(int i = 0; i < v.size() && v[i] <= m; i++){
		if(v[i] < n) continue;
		if(check(v[i])) cout << v[i] << endl;//如果是质数就输出
	}

	return 0;
}

时间复杂度：$O(n\sqrt n)$