Escape

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

The princess is going to escape the dragon's cave, and she needs to plan it carefully.

The princess runs at vp miles per hour, and the dragon flies at vd miles per hour. The dragon will discover the escape after t hours and will chase the princess immediately. Looks like there's no chance to success, but the princess noticed that the dragon is very greedy and not too smart. To delay him, the princess decides to borrow a couple of bijous from his treasury. Once the dragon overtakes the princess, she will drop one bijou to distract him. In this case he will stop, pick up the item, return to the cave and spend f hours to straighten the things out in the treasury. Only after this will he resume the chase again from the very beginning.

The princess is going to run on the straight. The distance between the cave and the king's castle she's aiming for is c miles. How many bijous will she need to take from the treasury to be able to reach the castle? If the dragon overtakes the princess at exactly the same moment she has reached the castle, we assume that she reached the castle before the dragon reached her, and doesn't need an extra bijou to hold him off.

Input

The input data contains integers vp, vd, t, f and c, one per line (1 ≤ vp, vd ≤ 100, 1 ≤ t, f ≤ 10, 1 ≤ c ≤ 1000).

Output

Output the minimal number of bijous required for the escape to succeed.

Sample Input

Input
1
2
1
1
10
Output
2
Input
1
2
1
1
8
Output
1

Hint

In the first case one hour after the escape the dragon will discover it, and the princess will be 1 mile away from the cave. In two hours the dragon will overtake the princess 2 miles away from the cave, and she will need to drop the first bijou. Return to the cave and fixing the treasury will take the dragon two more hours; meanwhile the princess will be 4 miles away from the cave. Next time the dragon will overtake the princess 8 miles away from the cave, and she will need the second bijou, but after this she will reach the castle without any further trouble.

The second case is similar to the first one, but the second time the dragon overtakes the princess when she has reached the castle, and she won't need the second bijou.

 #include<stdio.h>
#include<string.h>
int main()
{
double Vp,Vd,t,f,c;
while(scanf("%lf %lf %lf %lf %lf",&Vp,&Vd,&t,&f,&c)!=EOF)
{
float s;
int n=;
if(Vp>=Vd)
n=;
else
{
s=((Vp*t/(Vd-Vp))*Vd);
while(s<c)
{
n++;
s=((s+(s/Vd+f)*Vp)/(Vd-Vp))*Vd;
}
}
printf("%d\n",n);
}
return ;
}

CodeForces 148B Escape的更多相关文章

  1. Codeforces 148B: Escape

    题目链接:http://codeforces.com/problemset/problem/148/B 题意:公主从龙的洞穴中逃跑,公主的速度为vp,龙的速度为vd,在公主逃跑时间t时,龙发现公主逃跑 ...

  2. @codeforces - 932F@ Escape Through Leaf

    目录 @description@ @solution@ @accepted code@ @details@ @description@ 给定一个 n 个点的树(标号1~n),以结点 1 为根.每个结点 ...

  3. 寒假训练3解题报告 CodeForces #148

    CodeForces 148B 一道简单模拟,判断龙能够抓到公主几次,如果公主和龙同时到达公主的城堡,不算龙抓住她,因为路程除以速度可能会产生浮点数,所以这里考虑一下精度问题 #include < ...

  4. 水题 Codeforces Round #105 (Div. 2) B. Escape

    题目传送门 /* 水题:这题唯一要注意的是要用double,princess可能在一个小时之内被dragon赶上 */ #include <cstdio> #include <alg ...

  5. Codeforces 932.F Escape Through Leaf

    F. Escape Through Leaf time limit per test 3 seconds memory limit per test 256 megabytes input stand ...

  6. Codeforces Round #463 F. Escape Through Leaf (李超线段树合并)

    听说正解是啥 set启发式合并+维护凸包+二分 根本不会啊 , 只会 李超线段树合并 啦 ... 题意 给你一颗有 \(n\) 个点的树 , 每个节点有两个权值 \(a_i, b_i\) . 从 \( ...

  7. ICM Technex 2017 and Codeforces Round #400 (Div. 1 + Div. 2, combined) D. The Door Problem 2-SAT

    题目链接:http://codeforces.com/contest/776/problem/D D. The Door Problem time limit per test 2 seconds m ...

  8. Codeforces Round #287 (Div. 2) E. Breaking Good 最短路

    题目链接: http://codeforces.com/problemset/problem/507/E E. Breaking Good time limit per test2 secondsme ...

  9. [CCPC2019秦皇岛] E. Escape

    [CCPC2019秦皇岛E] Escape Link https://codeforces.com/gym/102361/problem/E Solution 观察到性质若干然后建图跑最大流即可. 我 ...

随机推荐

  1. C# PDFBox 解析PDF文件

    下载 PDFBox-0.7.3.zip PDFBox-0.7.3.dlllucene-demos-2.0.0.dlllucene-core-2.0.0.dllbcmail-jdk14-132.dllb ...

  2. 如何通过类找到对应的jar包

    ctrl+shift+T 然后输入对应类  

  3. YUI Reset CSS (学习摘抄)

    正在使用CSS的你,用过CSS Reset吗?当然,或许你用了,却不知道正在用,比如你可能用到: *{    margin: 0;    border: 0;    padding: 0;   } 这 ...

  4. android——单点触控移动,多点触控放大缩小

    xml <RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android" xmlns:to ...

  5. 【python cookbook】【字符串与文本】7.定义实现最短匹配的正则表达式

    问题:使用正则表达式对文本模式匹配,将识别出来的最长的可能匹配修改为找出最短的可能匹配 解决方法:在匹配模式中的*操作符后加上?修饰符 import re # Sample text text = ' ...

  6. blade模版之页面的嵌套

    blade模版 相关关键词:@section @yield @extends @extends @show @parent(追加内容而不是覆盖) 父页面view\layout\f.blade.php ...

  7. CentOS下使用Percona XtraBackup对MySQL5.6数据库innodb和myisam的方法

    Mysql卸载从下往上顺序 [root@localhost /]# rpm -e --nodeps qt-mysql-4.6.2-26.el6_4.x86_64[root@localhost /]# ...

  8. XCode属性面板使用说明

    Xcode 中Interface Builder 工具 是一个功能强大的“所见即所得”开发工具.本文主要介绍属性面板 和  对象库面板 对象库面板: 提供了所有Cocoa Touch 库给我们定义好的 ...

  9. MySQL存储引擎中的MyISAM和InnoDB区别详解

    在使用MySQL的过程中对MyISAM和InnoDB这两个概念存在了些疑问,到底两者引擎有何分别一直是存在我心中的疑问.为了解开这个谜题,搜寻了网络,找到了如下信息: MyISAM是MySQL的默认数 ...

  10. Python升级Yum不能使用解决

    1.系统版本 [root@vm10-254-206-95 ~]# cat /etc/issue CentOS release 6.4 (Final) Kernel \r on an \m 2.系统默认 ...