A. Music
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

题意：Lesha在线听歌， 但是网速不够快 每过q秒只能下载q - 1 秒， 这首歌总共t秒， 一开始已经下载了s秒， 每次从头播放， 每当播放到未下载的地方， 就会重新播放， 问一共播放几次？
解：
设第一次 经过x秒 听得时间和下载的时间相同 即s + (q - 1) / q * x = x（要把这个弯转过来） 解得 x = q * s
设第二次 经过t秒 听得时间和下载的时间相同 即x + (q - 1) / q * x = x （注意是x 而不是 s + x 看来是我老了傻了） 解得 t = q * x
这就看出来咯 每一次 听得时间和下载时间相同时 就是原起始时间乘q 然后边界情况自己想呗。

ac代码：

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 5;

int main(){
    ll t, s, q, sum = 0;
    while(cin >> t >> s >> q){ // 为啥用多组输入呢， 因为输入测试数据的时候方便
        ll x;
        sum = 1;
        x = q * s;
        while(x < t){
            sum ++;
            x = q * x;
        }
        cout << sum << "\n";
    }
    return 0;
}