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
- Java高效编程之二【对所有对象都通用的方法】
对于所有对象都通用的方法,即Object类的所有非final方法(equals.hashCode.toString.clone和finalize)都有明确的通用约定,都是为了要被改写(override ...
- hibernate笔记02
- IE11打不开网页, 所有菜单都被禁用了。
估计是安装完PPS之后,PPS安装程序附加了一些加载项到浏览器,而我在安装时强制禁用了它的加载项引起的. 解决方法是重置IE设置,命令为:inetcpl.cpl,点击高级选项卡的重置即可.
- wordpress网站被挂马以及防御方法
wordpress本身的安全性是非常的高的,一般不会被轻易的破解,被挂马,但是我们也不能够过度迷信wordpress的安全性,凡是连接上互联网的服务器和电脑,都存在被破解的风险性.所以我们在日常维护自 ...
- JavaEE基础(十五)/集合
1.集合框架(对象数组的概述和使用) A:案例演示 需求:我有5个学生,请把这个5个学生的信息存储到数组中,并遍历数组,获取得到每一个学生信息. Student[] arr = new Student ...
- 还原数据库,恢复SQLSERVER登录名的问题
还原SQLSERVER数据库,原来的数据库的于当前SQLSERVER同名用户就不能再登录了,原因是当前SQLSERVERD的master数据库的sysxlogins表的的sid与还原后的数据库的sys ...
- pipe row的用法, Oracle split 函数写法.
为了让 PL/SQL 函数返回数据的多个行,必须通过返回一个 REF CURSOR 或一个数据集合来完成.REF CURSOR 的这种情况局限于可以从查询中选择的数据,而整个集合在可以返回前,必须进行 ...
- 收集 关于php的博文
1. 小狼的世界: 浅谈用php实现mvc:http://www.cnblogs.com/cocowool/archive/2009/09/08/1562874.html 关于MVC的定义和解释,可以 ...
- eclipse+maven 无法编译
Archive for required library: 'F:/mavenLib/org/mybatis/mybatis/3.4.1/mybatis-3.4.1.jar' in project ' ...
- SQL数据类型大全 《转自网络》
数据类型是数据的一种属性,表示数据所表示信息的类型.任何一种计算机语言都定义了自己的数据类型.当然,不同的程序语言都具有不同的特点,所定义的数据类型的种类和名称都或多或少有些不同.SQLServer ...