Codeforces Round #315 (Div. 1) A. Primes or Palindromes? 暴力

A. Primes or Palindromes?
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://poj.org/problem?id=3261

Description

Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable.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

1 1

Sample Output

40

HINT

 

题意

给你p,q，A=p/q

π(n)表示1-n中素数的个数

rub(n)表示1-n中回文数的个数

求最大的n，满足π(n) ≤ A·rub(n).

题解：

CF测评姬很快，1e7直接上暴力就好了

暴力算出极限数据大概是1.5*1e6的样子，所以1e7很稳

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 10050000
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************
bool flag[maxn];
int primes[maxn], pi;
int pa[maxn];
void GetPrime()
{
    int i, j;
    pi = 0;
    memset(flag, false, sizeof(flag));
    for (i = 2; i < maxn; i++)
    {
        primes[i]+=primes[i-1];
        if (!flag[i])
        {
            primes[i] ++ ;
            for (j = i; j < maxn; j += i)
                flag[j] = true;
        }
    }
}
int check_Pa(int x)
{
    int X=x;
    int x2=0;
    while(X)
    {
        x2*=10;
        x2+=X%10;
        X/=10;
    }
    return x2==x?1:0;
}
void GetPa()
{
    for(int i=1;i<maxn;i++)
    {
        pa[i]+=pa[i-1];
        if(check_Pa(i))
            pa[i]++;
    }
}
int main()
{
    GetPrime();
    GetPa();
    double p,q;
    cin>>p>>q;
    double A=p/q;
    int ans=0;
    for(int i=1;i<maxn;i++)
    {
        if(A*pa[i]*1.0>=primes[i]*1.0)
            ans=i;
    }
    if(ans==0)
        cout<<"Palindromic tree is better than splay tree"<<endl;
    else
        cout<<ans<<endl;
}