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.

Examples

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.

【题目链接】:http://codeforces.com/contest/569/problem/A

【题解】



其实可以看成两辆车的追击问题;

一辆车A一开始在起点,另一辆B在S位置;

vA=1,vB=(q-1)/q;

则设经过了时间t

sa=t;

sb=s+(q-1)/q *t

令sa=sb

->t = q*s;

如果sb< t则sa又变回0,ans递增,此时s变成s+(q-1)/1*t;

然后重复上述步骤即可;

有点恶心。



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x) typedef pair<int,int> pii;
typedef pair<LL,LL> pll; //const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0); int t,s,q; int main()
{
//freopen("F:\\rush.txt","r",stdin);
rei(t);rei(s);rei(q);
int now = s;
int bo = 0;
int ans = 1;
while (now<t)
{
//v download ->q/q-1
int tempt = q*s;
int d = (q-1)*(tempt/q);
now+=d;
if (now >=t) break;
ans++;
s = now;
}
printf("%d\n",ans);
return 0;
}

【30.49%】【codeforces 569A】Music的更多相关文章

  1. JAVA 基础编程练习题49 【程序 49 子串出现的个数】

    49 [程序 49 子串出现的个数] 题目:计算字符串中子串出现的次数 package cskaoyan; public class cskaoyan49 { public static void m ...

  2. 【 BowWow and the Timetable CodeForces - 1204A 】【思维】

    题目链接 可以发现 十进制4 对应 二进制100 十进制16 对应 二进制10000 十进制64 对应 二进制1000000 可以发现每多两个零,4的次幂就增加1. 用string读入题目给定的二进制 ...

  3. 【52.49%】【codeforces 556A】Case of the Zeros and Ones

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  4. 【30.93%】【codeforces 558E】A Simple Task

    time limit per test5 seconds memory limit per test512 megabytes inputstandard input outputstandard o ...

  5. 【30.36%】【codeforces 740D】Alyona and a tree

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

  6. 【30.23%】【codeforces 552C】Vanya and Scales

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  7. 【59.49%】【codeforces 554B】Ohana Cleans Up

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

  8. 【30.43%】【codeforces 746C】Tram

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  9. 【codeforces 776E】The Holmes Children

    [题目链接]:http://codeforces.com/contest/776/problem/E [题意] f(n)是小于n的不同整数对(x,y)这里x+y==n且gcd(x,y)==1的个数; ...

随机推荐

  1. 11lession-class 类

    python既然也是面向对象编程的语言,自然也就跟java相似,它也有类的概念.今天就简单学习下.看如下代码 #!/usr/bin/python class cl_test: test = 0 def ...

  2. android图片特效处理之光晕效果

    这篇将讲到图片特效处理的图片光晕效果.跟前面一样是对像素点进行处理,本篇实现的思路可参见android图像处理系列之九--图片特效处理之二-模糊效果和android图像处理系列之十三--图片特效处理之 ...

  3. 65.十一级指针实现百万qq号的增删查改以及排序写入

    运行结果: 内存使用情况: 写入文件排序好的数据: 创建文件地址以及创建十一级指针 char *path = "QQ.txt"; char *sortpath = "QQ ...

  4. Flume的client

    Client:生产数据,运行在一个独立的线程.

  5. go语言函数作为参数传递

    go语言函数作为参数传递,目前给我的感觉几乎和C/C++一致.非常的灵活. import "fmt" import "time" func goFunc1(f ...

  6. 一个统一将数据转换为JSON的方法

    这是我得方法: 导包: import net.sf.json.JSONArray; import net.sf.json.JSONObject; public void writeJson(Objec ...

  7. JS错误记录 - To-do List

    var data = (localStorage.getItem('todolist'))? JSON.parse(localStorage.getItem('todolist')) : { todo ...

  8. BZOJ2160: 拉拉队排练(Manacher)

    Description 艾利斯顿商学院篮球队要参加一年一度的市篮球比赛了.拉拉队是篮球比赛的一个看点,好的拉拉队往往能帮助球队增加士气,赢得最终的比赛.所以作为拉拉队队长的楚雨荨同学知道,帮助篮球队训 ...

  9. MYSQLMANAGER实例管理器总结

    好久没有写文章了,今天来看看MYSQL的实例管理器(MYSQLMANAGER).一.简单介绍:1.MySQL实例管理器(IM)是通过TCP/IP端口运行的后台程序,用来监视和管理MySQL数据库服务器 ...

  10. HDU——T 1576 A/B

    http://acm.hdu.edu.cn/showproblem.php?pid=1576 Time Limit: 1000/1000 MS (Java/Others)    Memory Limi ...