C. Primes or Palindromes?
time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable.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

Examples
input
1 1
output
40
input
1 42
output
1
input
6 4
output
172
题意:不大于n的素数的个数<=A*不大于n的回文数的个数;求最大的n
思路:暴力;
#include<bits/stdc++.h>
using namespace std;
int a[];
int check(int x)
{
int ji=;
while(x)
{
a[ji++]=x%;
x/=;
}
for(int i=;i<ji/;i++)
{
if(a[i]!=a[ji--i])
return ;
}
return ;
}
int prime(int n)
{
if(n<=)
return ;
if(n==)
return ;
if(n%==)
return ;
int k, upperBound=n/;
for(k=; k<=upperBound; k+=)
{
upperBound=n/k;
if(n%k==)
return ;
}
return ;
}
#define ll __int64
const int N=2e6+;
ll p[N],h[N];
int main()
{
for(ll i=;i<=;i++)
{
p[i]=p[i-];
h[i]=h[i-];
if(prime(i))
p[i]++;
if(check(i))
h[i]++;
}
ll u,v;
scanf("%I64d%I64d",&u,&v);
for(int i=;i>;i--)
if(v*p[i]<=u*h[i])
{
printf("%d\n",i);
return ;
}
printf("Palindromic tree is better than splay tree\n");
return ;
}

Codeforces Round #315 (Div. 2) C. Primes or Palindromes? 暴力的更多相关文章

  1. 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 ...

  2. Codeforces Round #315 (Div. 2C) 568A Primes or Palindromes? 素数打表+暴力

    题目:Click here 题意:π(n)表示不大于n的素数个数,rub(n)表示不大于n的回文数个数,求最大n,满足π(n) ≤ A·rub(n).A=p/q; 分析:由于这个题A是给定范围的,所以 ...

  3. Codeforces Round #315 (Div. 2)——C. Primes or Palindromes?

    这道题居然是一个大暴力... 题意: π(n):小于等于n的数中素数的个数 rub(n) :小于等于n的数中属于回文数的个数 然后给你两个数p,q,当中A=p/q. 然后要你找到对于给定的A.找到使得 ...

  4. Codeforces Round #297 (Div. 2)D. Arthur and Walls 暴力搜索

    Codeforces Round #297 (Div. 2)D. Arthur and Walls Time Limit: 2 Sec  Memory Limit: 512 MBSubmit: xxx ...

  5. Codeforces Round #315 (Div. 2) (ABCD题解)

    比赛链接:http://codeforces.com/contest/569 A. Music time limit per test:2 seconds memory limit per test: ...

  6. codeforces 568a//Primes or Palindromes?// Codeforces Round #315 (Div. 1)

    题意:求使pi(n)*q<=rub(n)*p成立的最大的n. 先收集所有的质数和回文数.质数好搜集.回文数奇回文就0-9的数字,然后在头尾添加一个数.在x前后加a,就是x*10+a+a*pow( ...

  7. Codeforces Round #589 (Div. 2) C - Primes and Multiplication(数学, 质数)

    链接: https://codeforces.com/contest/1228/problem/C 题意: Let's introduce some definitions that will be ...

  8. Codeforces Round #315 (Div. 2)

    这次可以说是最糟糕的一次比赛了吧, 心没有静下来好好的去思考, 导致没有做好能做的题. Problem_A: 题意: 你要听一首时长为T秒的歌曲, 你点击播放时会立刻下载好S秒, 当你听到没有加载到的 ...

  9. Codeforces Round #315 (Div. 2B) 569B Inventory 贪心

    题目:Click here 题意:给你n,然后n个数,n个数中可能重复,可能不是1到n中的数.然后你用最少的改变数,让这个序列包含1到n所有数,并输出最后的序列. 分析:贪心. #include &l ...

随机推荐

  1. monggodb 模糊查询

    MongoDB的模糊查询其实很简单:      11.LIKE模糊查询userName包含A字母的数据(%A%)       SQL:SELECT * FROM UserInfo WHERE user ...

  2. php学习之有用的资源 总结

    1.php之道,http://laravel-china.github.io/php-the-right-way/ 2.改变php变成效率 http://www.php100.com/html/duj ...

  3. 修改hosts文件不起作用

    今天遇到个很奇怪的问题,在hosts文件里添加了一些域名指向后,发现根本没起作用,后来还发现个细节,就是hosts文件左下角有个小锁的标志,开始以为是杀毒软件之类的把他锁了.可是没找到在哪里有相关操作 ...

  4. 模块 - json/pickle/shelve/xml/configparser

    序列化: 序列化是指把内存里的数据类型转变成字符串,以使其能存储到硬盘或通过网络传输到远程,因为硬盘或网络传输时只能接受bytes. 为什么要序列化: 有种办法可以直接把内存数据(eg:10个列表,3 ...

  5. Android项目使用Ant多渠道打包(最新sdk)

    参考文章: http://blog.csdn.net/liuhe688/article/details/6679879 http://www.eoeandroid.com/thread-323111- ...

  6. python 脚本转成exe可执行程序

    本文所使用的工具是cx_Freeze,相对py2exe和PyInstaller来说,cx_Freeze的兼容性更好,支持最新的Python 3.X,支持PyQT,并且可以跨平台支持Windows和Li ...

  7. Rest Framework源码流程解析

    目录: 一 整体流程 二 具体流程 一 请求进来之后,都要先执行dispatch方法,dispatch方法根据请求方式的不同触发get/post/put/delete等方法 注意,APIView中的d ...

  8. arcgis中给属性文件加x y坐标

    两种方式: 一, 1在ArcGIS 9.2桌面软件arcview级别以上软件中,加载要添加x,y坐标的数据,打开属性表,添加X.Y字段 2 右键X字段,选择calculate geometry,如果颜 ...

  9. springmvc ModelAndView

    /** * 目标方法的返回值可以是 ModelAndView 类型. * 其中可以包含视图和模型信息 * SpringMVC 会把 ModelAndView 的 model 中数据放入到 reques ...

  10. Sublime Text Shortcuts

    Keyboard Shortcuts - Windows/Linux Warning This topic is a draft and may contain wrong information. ...