A. Eleven
time limit per test :

1 second

memory limit per test: 

256 megabytes

input: 

standard input

output: 

standard output

Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to have exactly n characters.

Her friend suggested that her name should only consist of uppercase and lowercase letters 'O'. More precisely, they suggested that the i-th letter of her name should be 'O' (uppercase) if i is a member of Fibonacci sequence, and 'o' (lowercase) otherwise. The letters in the name are numbered from 1 to n. Fibonacci sequence is the sequence f where

  • f1 = 1,
  • f2 = 1,
  • fn = fn - 2 + fn - 1 (n > 2).

As her friends are too young to know what Fibonacci sequence is, they asked you to help Eleven determine her new name.

Input

The first and only line of input contains an integer n (1 ≤ n ≤ 1000).

Output

Print Eleven's new name on the first and only line of output.

Examples
input
8
output
OOOoOooO
input
15
output
OOOoOooOooooOoo
 #include<bits/stdc++.h>
using namespace std; int FB[]; void init()
{
int t;
for(int i = ; i < ; i++)
FB[i] = ;
FB[] = FB[] = ;
for(int i = ; i < ;i++)
{ FB[i] = FB[i-] + FB[i-];
}
} int main()
{
init();
int n; while(~scanf("%d",&n))
{
int temp = ;
for(int i = ; i <= n; i++)
{
if(i == FB[temp])
{
printf("O");
temp++;
}
else
{
printf("o");
}
}
printf("\n");
}
return ;
}

Eleven的更多相关文章

  1. Eleven puzzle_hdu_3095(双向广搜).java

    Eleven puzzle Time Limit: 20000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) T ...

  2. Codeforces Round #459 (Div. 2)-A. Eleven

    A. Eleven time limit per test1 second memory limit per test256 megabytes Problem Description Eleven ...

  3. B - Eleven

    Problem description Eleven wants to choose a new name for herself. As a bunch of geeks, her friends ...

  4. python爬虫-携程-eleven参数

    携程-eleven分析 一.eleven的位置 通过对旁边栈的分析,它是在另一个js文件中调用的.那个js文件是一个自调用的函数,所以我们可以直接copy下来,用浏览器执行看看 执行运行是会报错的,u ...

  5. UVA 12672 Eleven(DP)

    12672 - Eleven Time limit: 5.000 seconds In this problem, we refer to the digits of a positive integ ...

  6. TIJ——Chapter Eleven:Holding Your Objects

    Java Provides a number of ways to hold objects: An array associates numerical indexes to objects. It ...

  7. Eleven scrum meeting 2015/11/5

    今日工作情况 小组成员 今日完成的工作 明日待做任务 唐彬 选课和退课模块 测试 赖彦谕 病情较重,请假 病情较重,请假 金哉仁 设计app logo 测试 闫昊 调整课程简介的展示效果 整合各个模块 ...

  8. 100-days: eleven

    Title: Facebook's live streaming(网络直播) is criticized(批评) after mosque(清真寺) shooting(枪击). live adj.现场 ...

  9. UVALive 6529 Eleven 区间dp

    题目链接:option=com_onlinejudge&Itemid=8&page=show_problem&problem=4540">点击打开链接 题意: ...

随机推荐

  1. 阿里云下Linux服务器安装Mysql、mongodb

    阿里云下Linux服务器安装Mysql.mongodb 一.MySQL的安装和配置 1.安装rpm包 rpm -Uvh http://dev.mysql.com/get/mysql-community ...

  2. angularJs-route路由详解

    本篇基于ng-route来讲下angular中的路由,路由功能主要是 $routeProvider服务 与 ng-view 实现. ng-view的实现原理,是根据路由的切换,动态编译html模板-- ...

  3. [LeetCode] Longest Line of Consecutive One in Matrix 矩阵中最长的连续1

    Given a 01 matrix M, find the longest line of consecutive one in the matrix. The line could be horiz ...

  4. 使用python实现人脸检测

    人脸检测 人脸检测使用到的技术是OpenCV,上一节已经介绍了OpenCV的环境安装,点击查看. 功能展示 识别一种图上的所有人的脸,并且标出人脸的位置,画出人眼以及嘴的位置,展示效果图如下: 多张脸 ...

  5. servlet之session设置

    商品对象,购物车对象,servlet的实现 商品: package app02d;public class Product {    private int id;    private String ...

  6. [TJOI2017]可乐

    题目描述 加里敦星球的人们特别喜欢喝可乐.因而,他们的敌对星球研发出了一个可乐机器人,并且放在了加里敦星球的1号城市上.这个可乐机器人有三种行为: 停在原地,去下一个相邻的城市,自爆.它每一秒都会随机 ...

  7. Linux LCD 显示图片【转】

    转自:https://blog.csdn.net/niepangu/article/details/50528190 BMP和JPEG图形显示程序1)  在LCD上显示BMP或JPEG图片的主流程图首 ...

  8. django rest-framework 2.请求和响应

    一.请求对象 REST 框架引入Request来扩展常规的HttpRequest,并提供了更灵活的请求解析.Request对象的核心功能是request.data属性. 导入方式: from rest ...

  9. Windows提示dll组件丢失

    我们在运行一些软件时,常常会遇到这种问题.下面就来提供解决办法: 登陆网址:www.dll-files.com. 找到页面的搜索部分,并且进行相关搜索: 下图显示了相关的dll下载链接. 下载解压即可 ...

  10. Python中高阶函数讲解

    高阶函数讲解 1. 常规高阶函数 递归函数 格式:def func_name(variable): '''__doc__'''#函数的说明文档 if 条件表达式:#限制递归退出值 pass retur ...