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

探讨了素数与回文数的数量对比问题，提出了一个算法来寻找满足特定条件的最大整数n，使得n及其以下的素数数量不超过回文数数量乘以系数A。

Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable.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 ( $\text{[math]}$ , $\text{[math]}$ ).

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;
}