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Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable.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
,
).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[];
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//**************************************************************************************
bool flag[maxn];
int primes[maxn], pi;
int pa[maxn];
void GetPrime()
{
int i, j;
pi = ;
memset(flag, false, sizeof(flag));
for (i = ; i < maxn; i++)
{
primes[i]+=primes[i-];
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=;
while(X)
{
x2*=;
x2+=X%;
X/=;
}
return x2==x?:;
}
void GetPa()
{
for(int i=;i<maxn;i++)
{
pa[i]+=pa[i-];
if(check_Pa(i))
pa[i]++;
}
}
int main()
{
GetPrime();
GetPa();
double p,q;
cin>>p>>q;
double A=p/q;
int ans=;
for(int i=;i<maxn;i++)
{
if(A*pa[i]*1.0>=primes[i]*1.0)
ans=i;
}
if(ans==)
cout<<"Palindromic tree is better than splay tree"<<endl;
else
cout<<ans<<endl;
}
Codeforces Round #315 (Div. 1) A. Primes or Palindromes? 暴力的更多相关文章
- Codeforces Round #315 (Div. 2) C. Primes or Palindromes? 暴力
C. Primes or Palindromes? time limit per test 3 seconds memory limit per test 256 megabytes input st ...
- 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 ...
- 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 ...
- 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 ...
- 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: ...
- 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( ...
- 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 ...
- 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 ...
- 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
- 移动对meta的定义
以下是meta每个属性详解 尤其要注意的是content里多个属性的设置一定要用分号+空格来隔开,如果不规范将不会起作用. 一.<meta http-equiv="Content-Ty ...
- android直接读取数据库文件
public class Dictionary extends Activity implements OnClickListener, TextWatcher{ private final ...
- HDU 5114 Collision
Collision Time Limit: 15000/15000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others) Total ...
- Webdriver API (三)- actions
Actions类主要定义了一些模拟用户的鼠标mouse,键盘keyboard操作.对于这些操作,使用perform()方法进行执行. actions类可以完成单一的操作,也可以完成几个操作的组合. 有 ...
- iOS多线程之GCD小记
iOS多线程之GCD小记 iOS多线程方案简介 从各种资料中了解到,iOS中目前有4套多线程的方案,分别是下列4中: 1.Pthreads 这是一套可以在很多操作系统上通用的多线程API,是基于C语言 ...
- SQL对字符串进行排序
假设字符串中只由'A'.'B'.'C'.'D'组成,且长度为7.并设函数REPLICATE(<字符串>,<n>)可以创建一个<字符串>的n个副本的字符串,另外还有R ...
- 机器学习真的可以起作用吗?(3)(以二维PLA为例)
前两篇文章已经完成了大部分的工作,这篇文章主要是讲VC bound和 VC dimension这两个概念. (一)前文的一点补充 根据前面的讨论,我们似乎只需要用来替代来源的M就可以了,但是实际公式却 ...
- poj2396 Budget(有源汇上下界可行流)
[题目链接] http://poj.org/problem?id=2396 [题意] 知道一个矩阵的行列和,且知道一些格子的限制条件,问一个可行的方案. [思路] 设行为X点,列为Y点,构图:连边(s ...
- CSS Sprite的优缺点分析
目前大多数的开发人员对这个技术都有相当地掌握,也有很多关于它的教程和文章.几乎所有的文章中都宣称设计师和开发人员都应该使用 CSS sprite 来减少 HTTP 请求数,并且节省一些流量.这个技术被 ...
- Hadoop 删除节点步骤
1.在hadoop1.1.1/conf 下新建文件 nn-excluded-list 并写入要删除的节点名称或者IP 一个节点 一行 如: mos5200app cmpaknwom rac7 2.分发 ...