Codeforces 568 A Primes or Palindromes?（求素数个数+判断是否是回文数）

最新推荐文章于 2025-05-24 15:58:27 发布

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最新推荐文章于 2025-05-24 15:58:27 发布

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Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable.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

没什么好说的当时做的时候以为关系有单调性还以为要用二分。。其实数到13000000就达到那个42倍关系的极限了所以直接从大的开始暴力枚举就能求出答案另外不存在没有答案的情况

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 1300000
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int Read()
{
    char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    int x = 0;
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

void Print(int a)
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}

int isPalindrome(int x)
{
  //if (x < 0) return false;
  int div = 1;
  while (x / div >= 10) {
    div *= 10;
  }
  while (x != 0) {
    int l = x / div;
    int r = x % 10;
    if (l != r) return 0;
    x = (x % div) / 10;
    div /= 100;
  }
  return 1;
}  //回文数的判断

int pri[MAXN];
void init()
{
    for(int i=0;i<MAXN;i++)
        pri[i]=1;
    pri[0]=pri[1]=0;
    for(int i=2;i<MAXN;i++)
    {
        if(pri[i])
        {
            if(i>MAXN/i) continue;
            for(int j=i*i;j<MAXN;j+=i)
                pri[j]=0;
        }
    }
    for(int i=1;i<MAXN;i++)
        pri[i]+=pri[i-1];
//    for(int i=1;i<=100;i++)
//        cout<<"i  "<<i<<"  "<<pri[i]<<endl;
}
int a[MAXN];

int main()
{
    //fread;
    init();
    MEM(a,0);
    for(int i=1;i<MAXN;i++)
        a[i]=isPalindrome(i);
    for(int i=1;i<MAXN;i++)
        a[i]+=a[i-1];
    int p,q;
    while(scanf("%d%d",&p,&q)!=EOF)
    {
        int ans;
        for(int i=MAXN-1;i>=1;i--)
        {
            if(q*pri[i]<=p*a[i])
            {
                ans=i;
                break;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}