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

2

Input

5 4 7

Output

1

Input

6 2 3

Output

1

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=0;
    cin>>t>>s>>q;
    while(s<t)
    {
       c++;
       s=s*q;
    }
    cout<<c<<endl;
    return 0;
 }