Music

Music

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88890#problem/C

题目：

Description

Little Lesha loves listening to music via his smartphone. But the smartphone doesn't have much memory, so Lesha listens to his favorite songs in a well-known social network InTalk.

Unfortunately, internet is not that fast in the city of Ekaterinozavodsk and the song takes a lot of time to download. But Lesha is quite impatient. The song's duration is T seconds. Lesha downloads the first S seconds of the song and plays it. When the playback reaches the point that has not yet been downloaded, Lesha immediately plays the song from the start (the loaded part of the song stays in his phone, and the download is continued from the same place), and it happens until the song is downloaded completely and Lesha listens to it to the end. For q seconds of real time the Internet allows you to download q - 1 seconds of the track.

Tell Lesha, for how many times he will start the song, including the very first start.

Input

The single line contains three integers T, S, q (2 ≤ q ≤ 10⁴, 1 ≤ S < T ≤ 10⁵).

Output

Print a single integer — the number of times the song will be restarted.

Sample Input

Input

5 2 2

Output

Input

5 4 7

Output

Input

6 2 3

Output

Hint

In the first test, the song is played twice faster than it is downloaded, which means that during four first seconds Lesha reaches the moment that has not been downloaded, and starts the song again. After another two seconds, the song is downloaded completely, and thus, Lesha starts the song twice.

In the second test, the song is almost downloaded, and Lesha will start it only once.

In the third sample test the download finishes and Lesha finishes listening at the same moment. Note that song isn't restarted in this case.

题意：

一首歌一共有T秒，s表示第一次下载的秒数，q表示每q秒可以下载(q-1)秒这首歌，第一次下载结束后，每次播放的同时会下载。求一共从头开始播放的次数。

分析：

一道很好的想法题，每当播放到之前的未download处时，由于播放的同时也在下载，所以总会再次下载了一些，理论上会无限循环播放和下载，但总会有个极限，可以记为S(当前可播放的时间)。

S=s+s*(q-1)/q+s*(q-1)/q*(q-1)/q+...。运用等比数列求和可以得知当n->+∞时，S=s*q;

当播放到第i次的时候S=s*(q^i)。

#include<iostream>

using namespace std;

int main()

{

    int t,s,q,c=;

    cin>>t>>s>>q;

    while(s<t)

    {

       c++;

       s=s*q;

    }

    cout<<c<<endl;

    return ;

 }

随机推荐

PS通过滤色实现简单的图片拼合
素材如下: 素材一: 雪山素材二: 月亮效果: 实现步骤 1.在PS中打开雪山素材一 2.将月亮素材直接拖入雪山所在的图层中 3.锁定置入素材的高宽比(点击一下链状按钮) 4.调整月亮到合适大 ...
Linux top命令排序
在系统维护的过程中,随时可能有需要查看 CPU 使用率,并根据相应信息分析系统状况的需要.在 Linux 中,可以通过 top 命令来查看 CPU 使用状况.运行 top 命令后,CPU 使用状态会以 ...
七牛portal可用性测试记
引言:2013年年底应七牛公司朋友的邀请,给他们的Portal进行可用性测试.七牛(http://www.qiniu.com/)一直专注于云存储基础服务,在业内颇有声誉.七牛云存储的后台选用并不常用的 ...
常用的PC/SC接口函数
PC/SC规范是一个基于WINDOWS平台的一个标准用户接口(API),提供了一个从个人电脑(Personal Computer)到智能卡(SmartCard)的整合环境,PC/SC规范建立在工业标准 ...
[Sciter系列] MFC下的Sciter&ndash&semi;4&period;HTML与图片资源内置
[Sciter系列] MFC下的Sciter–4.HTML与图片资源内置,防止代码泄露. 本系列文章的目的就是一步步构建出一个功能可用,接口基本完善的基于MFC框架的SciterFrame程序,以此作 ...
bzoj2257
不难发现其实这个捣鼓过程就是一个辗转相减所以最小燃料是瓶容量的最小公约数然后穷举约数即可 ..] of longint; n,k,x,i,m,j,t:longint; procedure sort(l ...
Intel hex 文件格式解密
Intel hex 文件常用来保存单片机或其他处理器的目标程序代码.它保存物理程序存储区中的目标代码映象.一般的编程器都支持这种格式. Intel hex 文件全部由可打印的ASCII字符组成(可以用 ...
Linux下获取线程TID的方法——gettid()
(转载)http://blog.csdn.net/delphiwcdj/article/details/8476547 如何获取进程的PID(process ID)? 可以使用: #include & ...
Numpy数组索引为-1和None
numpy的数组操作方便,可以用:来切片,用布尔数组或者布尔表达式来查找符合条件的数据,也可以用数组作为另一个数组的索引来查找指定的数据.但有时也会见到数组索引为-1和None.两者的用法如下: 1. ...
ICM Technex 2017 and Codeforces Round &num;400 (Div&period; 1 + Div&period; 2, combined)
前四题比较水,E我看出是欧拉函数傻逼题,但我傻逼不会,百度了下开始学,最后在加时的时候A掉了 AC:ABCDE Rank:182 Rating:2193+34->2227 终于橙了,不知道能待几 ...

随机推荐

相关文章