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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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; ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. AtCoder M-SOLUTIONS 2019 Task E. Product of Arithmetic Progression

    problem link Official editorial: code: int main() { #if defined LOCAL && !defined DUIPAI ifs ...

  2. springboot简易上传下载

    1.导入上传下载依赖: <dependency> <groupId>commons-fileupload</groupId> <artifactId>c ...

  3. 如何拿到美团offer的

    美团,我是在拉勾网上投的简历,之前也投过一次,简历都没通过删选,后来让学姐帮我改了一下简历,重新投另一个部门,获得了面试机会.10月23日,中午HR打电话过来预约了下午4点半面试,说会在线写代码,让我 ...

  4. Codeforces 1247D. Power Products

    传送门 要满足存在 $x$ ,使得 $a_i \cdot a_j = x^k$ 那么充分必要条件就是 $a_i \cdot a_j$ 质因数分解后每个质因数的次幂都要为 $k$ 的倍数 证明显然 设 ...

  5. Javascript中的继承与复用

    实现代码复用的方法包括:工厂模式.构造函数模式.原型模式(<高三>6.2章 P144),它们各自的特点归结如下:1.工厂模式虽然使创建对象一定程度上实现了代码复用,但却没有解决对象识别问题 ...

  6. vue + echarts 实现中国地图 展示城市

    Demo 安装依赖 vue中安装echarts npm install echarts -S 在main.js中引用 import echarts from 'echarts'Vue.prototyp ...

  7. 编写Dockerfile自定义镜像

    要求 编写一个Dockerfile自定义centos镜像,要求在容器内部可以使用vim和ifconfig命令,并且登入落脚点为/usr/local 编写Dockerfile FROM centos M ...

  8. 网络编程之NIO

    传统的BIO(Blocking IO)的缺点: 1.基于阻塞式IO建立起来的,导致服务端一直阻塞等待着客户端发起请求,如果客户端不发起,服务端的的业务线程会一直存. 2.弹性伸缩能力差,线程数和客户端 ...

  9. 搭建nginx环境

    1.安装nginx 下载地址:http://nginx.org/en/download.html 博主选择的是nginx1.8.1,点击下载 下载完成后是一个压缩包, 解压后双击nginx.exe 这 ...

  10. otool随笔测试

    otool 工具 查看库/反编译等二进制信息 1 依赖库查询 otool -L Payload/XXX.app/XXX 2 查看该应用是否砸壳 otool -l Payload/XXX.app/XXX ...