CodeForces 148B 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
1
2
1
1
10
2
1
2
1
1
8
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
- 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 ...
- @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 ...
- 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 < ...
- 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 ...
- Codeforces 932.F Escape Through Leaf
F. Escape Through Leaf time limit per test 3 seconds memory limit per test 256 megabytes input stand ...
- 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 \( ...
- 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 ...
- 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 ...
- [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
- 「Ruby && Sqlite3」How to install sqlite3 for ruby? (solve: sqlite-ruby no such file...)
error message: no such file .... 安装 gem install sqlite3-ruby -- --with-sqlite3-dir=/usr/lo ...
- Broadwell I7-5775c/5675c BSOD 蓝屏问题
今年6月末Intel发布了第五代台式机CPU,代号Broadwell,跳票一年多,仅有两款零售型号,I7-5775C和I5-5675C. 第一时间就买来试用,当时还是在日亚购买的,等待了10天左右终于 ...
- sql回滚
rollback是针对事务的,你如果没有在执行语句之前开启事务,那么无法rollback,建议你还是想别的办法吧,事务语句如下(sqlserver的给你借鉴):--开启事务begin tran --执 ...
- HDU 2236:无题II(二分搜索+二分匹配)
http://acm.hdu.edu.cn/showproblem.php?pid=2236 题意:中文题意. 思路:先找出最大和最小值,然后二分差值,对于每一个差值从下界开始枚举判断能不能二分匹配. ...
- MyEclipse+Android 安装配置
1.先安装M有Eclipse 就是不断点:下一步下一步最后finish 激活:http://blog.my-eclipse.cn/myeclipse-2014-crack.html (该网站上 ...
- [算法][C]计算向量的角度
C 语言里 double atan2(double y,double x) 返回的是原点至点(x,y)的方位角,即与 x 轴的夹角.也可以理解为复数 x+yi 的辐角.返回值的单位为弧度,取值范围为 ...
- 【转】身份证号码校验与信息提取 - Java 代码
转载地址:http://www.w3china.org/blog/more.asp?name=lhwork&id=19148 import java.util.regex.*; /** ...
- JS获取非行间样式
我们都知道用offset函数获取元素样式是一件很方便的事,但是offset只能获取行间样式,而无法获得非行间样式,这是它的瓶颈所在. 我们都知道js获取行间样式的方法,那么js是如何获取行距样式的呢? ...
- mybatis+springMVC新感悟
一直以为按照例子里写的.先编写User实体类,之后在编写User.xml之后在配置文件里指明接口文件.然后在controller中就可以通过就可以通过定义接口,在取值 IUserOperation u ...
- AreYouBusy
AreYouBusy Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Su ...