Codeforces Round #315 (Div. 2) C. Primes or Palindromes? (素数打表 回文数)

Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable.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 A find 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).

Sample test(s)

input

1 1

output

40

input

1 42

output

1

input

6 4

output

172

/* In the name of Allah */#include<bits/stdc++.h>using namespace std;const int max_n=1e6+2e5+5;int p[max_n];bool check[max_n];int q[max_n];int main(){    int a,b;    cin >> a>>b;    for(int i=2;i<max_n;i++)    {        if(!check[i])        {            p[i]=p[i-1]+1;            for(int j=2*i;j<max_n;j+=i)                check[j]=1;        }        else            p[i]=p[i-1];    }    for(int i=1;i<max_n;i++)    {        int id=i;        int j=0;        while(id!=0)        {            j=j*10 + id%10;            id/=10;        }        if(i==j) q[i]=q[i-1]+1;        else q[i]=q[i-1];    }    for(int i=max_n-5;i>0;i--)    {        if(b*p[i]<= a*q[i])        {            cout<< i;            return 0;        }    }}

大家所知的素数打表时间复杂度几乎都是n2。

void init_prime(){int i, j;for(i = 2;i <= sqrt(1000002.0); ++i){if(!prime[i])for(j = i * i; j < 1000002; j += i)prime[j] = 1;}j = 0;for(i = 2;i <= 1000002; ++i)if(!prime[i]) prime[j++] = i;}