题干：
Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable. A palindromic number is another matter. It is aesthetically pleasing, and it has a number of remarkable properties. Help Rikhail to convince the scientific community in this!
Let us remind you that a number is called prime if it is integer larger than one, and is not divisible by any positive integer other than itself and one.
Rikhail calls a number a palindromic if it is integer, positive, and its decimal representation without leading zeros is a palindrome, i.e. reads the same from left to right and right to left.
One problem with prime numbers is that there are too many of them. Let's introduce the following notation: π(n) — the number of primes no larger than n, rub(n) — the number of palindromic numbers no larger than n. Rikhail wants to prove that there are a lot more primes than palindromic ones.
He asked you to solve the following problem: for a given value of the coefficient Afind the maximum n, such that π(n) ≤ A·rub(n).
Input
The input consists of two positive integers p, q, the numerator and denominator of the fraction that is the value of A (,
).
Output
If such maximum number exists, then print it. Otherwise, print "Palindromic tree is better than splay tree" (without the quotes).
Examples
Input
1 1
Output
40
Input
1 42
Output
1
Input
6 4
Output
172题目大意：
定义π(N)=∑(i为素数?1:0)，i<=N， rub(N)=∑(i为回文数?1:0)，i<=N。A=p/q，求最大的N，使得π(n) ≤ A·rub(n)。
解题报告：
通过观察单调性（因为显然素数个数增长的快啦，可以打个表判断一下，n随便取个值，然后输出几组数据，就可以看出来，素数个数的增长远远大于回文数的增长，我这有个数据，100W个数，有1000个回文数左右），和p和q给的范围（给你范围的目的就是告诉你，可以打表？），我们知道可以算出一个极限来，可以算出 ，n是20W就足够了，数据量也不大，直接打表暴力。
AC代码：
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int MAX = 2e6+ 5;
int wei[1005],cnt;
ll su[1000005];
bool isprime[MAX];
ll sum1[MAX],sum2[MAX];//sum1素数，sum2回文数
void prime() {cnt = 0;memset(isprime,1,sizeof isprime);isprime[0] = isprime[1] =0;for(ll i = 2 ; i<MAX; i++) {if(isprime[i]) {su[++cnt] = i;}for(ll j = 1; i * su[j]< MAX && j <= cnt; j++) {isprime[i * su[j]] = 0;}}
}
bool fit(int x) {int p = 0;while(x) {wei[++p] = x%10;x/=10;}for(int i = 1; i<=(p+1)/2; i++) {if(wei[i] != wei[p-i+1]) return 0;}return 1;
}
int main()
{int n;//1000W 中 1W个回文数左右int p,q;scanf("%d%d",&p,&q);prime();for(int i = 1; i<MAX; i++) {sum1[i] = sum1[i-1];if(isprime[i]) sum1[i]++;}for(int i = 1; i<MAX; i++) {sum2[i] = sum2[i-1];if(fit(i)) sum2[i]++;}int ans = -1; for(int i = MAX-1; i>=0; i--) {if(sum1[i] * q <= sum2[i] * p) {ans = i;break;}}if(ans == -1) puts("Palindromic tree is better than splay tree");else printf("%d\n",ans);return 0 ;}
