cf 569A. Music (暴力)

A. Music

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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 test(s)

input

5 2 2

output

2

input

5 4 7

output

1

input

6 2 3

output

1

Note
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.

一开始写错了，原因是误以为时间是离散的。。。

/*************************************************************************
> File Name: code/cf/#315/A.cpp
> Author: 111qqz
> Email: rkz2013@126.com
> Created Time: 2015年08月11日 星期二 01时26分16秒
************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
int main()
{
int t,s,q;
cin>>t>>s>>q;
int ans = 1;
int x = s*q;
while (x<t)
{
ans++;
x = x*q;
}

cout<<ans<<endl;

return 0;
}

竟然整秒整秒地算。

然而实际上是连续的。

由于没过q (s),播放时间会比下载时间领先1s，那么初始的s(s),可以领先s*q （s），之后如果没放完就要重新开始放。这个时候又回到了之前的问题，只不过初始变成了s*q(s)。