A. 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input standard input output standard output Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and un…

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题目连接 http://poj.org/problem?id=3261 Description Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable.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 palindro…

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input standard input output standard output Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and un…

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,所以可以先暴力求下最大的n满足上式,可以想象下随着n的增大A也在增大(总体正相关,并不是严格递增的),所以二分查找时不行的,所以对给定的A,n是一定存在的.这个题的关键就是快速得到素数表最好在O(n)的时间以内.(杭电15多校的一个题也用到了这个算法点这里查看) #include <bits/stdc+…

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(10,2).偶回文同理.然后不能二分,因为比值不是单调的. 乱码: #pragma comment(linker,"/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include…

题目链接: 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input standard input output standard output Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex…

 Palindromes Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Description A regular palindrome is a string of numbers or letters that is the same forward as backward. For example, the string "ABCDEDCBA" is…

time limit per test3 seconds memory limit per test256 megabytes inputstandard input outputstandard output Rikhail Mubinchik believes that the current definition of prime numbers is obsolete as they are too complex and unpredictable. A palindromic num…

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π(n) ≤ A·rub(n) 最大的n. (A<=42) 思路: 首先我们能够暴力算出当n为大概150万左右的时候,π(n)大概是 rub(n) 的42倍. 所以我们仅仅须要for到150万左右就好,由于对于后面的式子.肯定能在150万的范围内找到一个n使得这个式子成立的. 并且.我们可以得出由于…

prime numbers non greater than n is about . We can also found the amount of palindrome numbers with fixed length k — it is about  which is . #include <iostream> #include <cstdio> #include <string.h> #include <cmath> using namespace…