Educational Codeforces Round 33 (Rated for Div. 2) B题
B. Beautiful Divisors
Recently Luba learned about a special kind of numbers that she calls beautiful numbers. The number is called beautiful iff its binary representation consists of k + 1 consecutive ones, and then k consecutive zeroes.
Some examples of beautiful numbers:
12 (110);
1102 (610);
11110002 (12010);
1111100002 (49610).
More formally, the number is beautiful iff there exists some positive integer k such that the number is equal to (2k - 1) * (2k - 1).
Luba has got an integer number n, and she wants to find its greatest beautiful divisor. Help her to find it!
Input
The only line of input contains one number n (1 ≤ n ≤ 105) — the number Luba has got.
Output
Output one number — the greatest beautiful divisor of Luba's number. It is obvious that the answer always exists.
Input
Output
思路:看了看数据范围。QAQ,发现可以打表===水过
AC代码:
#include<bits/stdc++.h> using namespace std;
int arr[]={,,, , ,, ,, ,};
int main(){
int n;
cin>>n;
for(int i=;i>=;i--){
if(arr[i]>n)
continue;
if(arr[i]==n){
printf("%d",n);return ;
}
if(n%arr[i]==){
cout<<arr[i];return ;
}
}
return ;
}
Educational Codeforces Round 33 (Rated for Div. 2) B题的更多相关文章
- Educational Codeforces Round 33 (Rated for Div. 2) D题 【贪心:前缀和+后缀最值好题】
D. Credit Card Recenlty Luba got a credit card and started to use it. Let's consider n consecutive d ...
- Educational Codeforces Round 33 (Rated for Div. 2) C题·(并查集变式)
C. Rumor Vova promised himself that he would never play computer games... But recently Firestorm — a ...
- Educational Codeforces Round 33 (Rated for Div. 2) A题
A. Chess For Three Alex, Bob and Carl will soon participate in a team chess tournament. Since they a ...
- Educational Codeforces Round 33 (Rated for Div. 2) E. Counting Arrays
题目链接 题意:给你两个数x,yx,yx,y,让你构造一些长为yyy的数列,让这个数列的累乘为xxx,输出方案数. 思路:考虑对xxx进行质因数分解,设某个质因子PiP_iPi的的幂为kkk,则这个 ...
- Educational Codeforces Round 33 (Rated for Div. 2) F. Subtree Minimum Query(主席树合并)
题意 给定一棵 \(n\) 个点的带点权树,以 \(1\) 为根, \(m\) 次询问,每次询问给出两个值 \(p, k\) ,求以下值: \(p\) 的子树中距离 \(p \le k\) 的所有点权 ...
- Educational Codeforces Round 33 (Rated for Div. 2) 题解
A.每个状态只有一种后续转移,判断每次转移是否都合法即可. #include <iostream> #include <cstdio> using namespace std; ...
- Educational Codeforces Round 33 (Rated for Div. 2)A-F
总的来说这套题还是很不错的,让我对主席树有了更深的了解 A:水题,模拟即可 #include<bits/stdc++.h> #define fi first #define se seco ...
- Educational Codeforces Round 33 (Rated for Div. 2) D. Credit Card
D. Credit Card time limit per test 2 seconds memory limit per test 256 megabytes input standard inpu ...
- Educational Codeforces Round 33 (Rated for Div. 2) C. Rumor【并查集+贪心/维护集合最小值】
C. Rumor time limit per test 2 seconds memory limit per test 256 megabytes input standard input outp ...
随机推荐
- log4j一些配置用法
Log4j基本用法----日志级别 基本使用方法: Log4j由三个重要的组件构成:日志信息的优先级,日志信息的输出目的地,日志信息的输出格式.日志信息的优先级从高到低有ERROR.WARN.INFO ...
- 最长回文 HDU - 3068(马拉车算法)
Problem Description 给出一个只由小写英文字符a,b,c...y,z组成的字符串S,求S中最长回文串的长度. 回文就是正反读都是一样的字符串,如aba, abba等 Input 输入 ...
- LC 387. First Unique Character in a String
题目描述 Given a string, find the first non-repeating character in it and return it's index. If it doesn ...
- dedecms发布文章时间显示多少分钟前
/**文章发布多少时间前*/function tranTime($time) { $rtime = date("m-d H:i",$time); $htime = date(&qu ...
- python基础(十)--函数进阶
嵌套函数 >>> graphic = '三角形' >>> def chang(): graphic = '正方形' def chang1(): #内部嵌套的函数命名 ...
- 社工工具包 SEToolkit
社会工程学(Social Engineering)简称社工,其通过分析攻击对象的心理弱点,利用人性的本能反应,以及任何好奇心,贪婪等心理特征进行的,使用诸如假冒,欺骗,引诱等多种手段来达成攻击目标的一 ...
- Python 变量作用域与函数
Python 的创始人为吉多·范罗苏姆(Guido van Rossum).1989年的圣诞节期间,吉多·范罗苏姆为了在阿姆斯特丹打发时间,决心开发一个新的脚本解释程序,作为ABC语言的一种继承.Py ...
- 怎样安装并编译TypeScript?
1. 使用: npm -v 查看是否安装了 npm , 如果没有安装, 请前往 Nodejs 官网 下载安装, 下图表示已经安装 npm , 版本为: 6.9.0 . PS C:\Users\Adm ...
- Scala学习二十一——隐式转换和隐式参数
一.本章要点 隐式转换用于类型之间的转换 必须引入隐式转换,并确保它们可以以单个标识符的形式出现在当前作用域 隐式参数列表会要求指定类型的对象.它们可以从当前作用域中以单个标识符定义的隐式对象的获取, ...
- Python的.sort()方法和sorted()比较总结
1,.sort()方法 使用方式是:列表.sort(),作用是将原来的列表正序排序,所以它是对原来的列表进行的操作,不会产生一个新列表,例如: import random numList=[] pri ...