杭电 1155 Bungee Jumping(物理题)
Problem Description
Unfortunately, he had not had
enough time to calculate whether the bungee rope has the right length, so it is
not clear at all what is going to happen when he jumps off the bridge. There are
three possible scenarios:
The rope is too short (or too strong), and James
Bond will never reach the ground.
The rope is too long (or too weak), and
James Bond will be going too fast when he touches the ground. Even for a special
agent, this can be very dangerous. You may assume that if he collides at a speed
of more than 10 m/s, he will not survive the impact.
The rope's length and
strength are good. James Bond touches the ground at a comfortable speed and can
escape.
As his employer, you would like to know whether James Bond survives
or whether you should place a job ad for the soon-to-be vacant position in the
local newspaper. Your physicists claim that:
The force with which James is
pulled towards the earth is
9.81 * w,
where w is his weight in kilograms
and 9.81 is the Earth acceleration in meters over squared seconds.
Mr. Bond
falls freely until the rope tautens. Then the force with which the bungee rope
pulls him back into the sky depends on the current length of the rope and is
k * Δl,
where Δl is the difference between the rope's current length and
its nominal, unexpanded length, and k is a rope-specific constant.
Given the
rope's strength k, the nominal length of the rope l in meters, the height of the
bridge s in meters, and James Bond's body weight w, you have to determine what
is going to happen to our hero. For all your calculations, you may assume that
James Bond is a point at the end of the rope and the rope has no mass. You may
further assume that k, l, s, and w are non-negative and that s <
200.
The input contains several test cases, one test case per line. Each
test case consists of four floating-point numbers (k, l, s, and w) that describe
the situation. Depending on what is going to happen, your program must print
"Stuck in the air.", "Killed by the impact.", or "James Bond survives.". Input
is terminated by a line containing four 0s, this line should not be processed.
Sample Input
Sample Output
/*
k 绳子劲度系数
l 绳长
s 桥的高度
w 体重
*/
#include<cstdio>
#define g 9.81
int main()
{
double k,l,s,w;
double Eg,Ek,Ev;
while(scanf("%lf %lf %lf %lf",&k,&l,&s,&w)&&(k+l+s+w))
{
Eg=w*g*s;
Ev=*w/;
Ek=k*(s-l)*(s-l)/;
if(l >= s)
{
if(Eg <= Ev)
printf("James Bond survives.\n");
else
printf("Killed by the impact.\n");
}
else
{
if(Eg < Ek)
printf("Stuck in the air.\n");
else if(Eg-Ek <= Ev)
printf("James Bond survives.\n");
else
printf("Killed by the impact.\n");
} } }
杭电 1155 Bungee Jumping(物理题)的更多相关文章
- HDU 1155 Bungee Jumping 物理
题目大意:给出k:绳子的劲度系数,l:绳长,s:桥高,w:邦德的质量,g取9.81.绳子弹力=形变量*劲度系数.如果落地速度大于10 则摔死,小于0则飘着空中. 题目思路:根据能量守恒得知:落地的动能 ...
- hdu 1155 Bungee Jumping
http://acm.hdu.edu.cn/showproblem.php?pid=1155 Bungee Jumping Time Limit: 2000/1000 MS (Java/Others) ...
- HDU 1155 Bungee Jumping(物理题,动能公式,弹性势能公式,重力势能公式)
传送门: Bungee Jumping Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Othe ...
- 杭电1087 Super Jumping! Jumping! Jumping!(初见DP)
Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 ...
- 胜利大逃亡(杭电hdu1253)bfs简单题
胜利大逃亡 Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Subm ...
- HDU 4968(杭电多校#9 1009题)Improving the GPA (瞎搞)
题目地址:HDU 4968 这题的做法是全部学科的学分情况枚举,然后推断在这样的情况下是否会符合平均分. 直接暴力枚举就可以. 代码例如以下: #include <cstring> #in ...
- 杭电1081 第二道 dfs题
Problem Description 呃......变形课上Harry碰到了一点小麻烦,因为他并不像Hermione那样能够记住所有的咒语而随意的将一个棒球变成刺猬什么的,但是他发现了变形咒语的一个 ...
- HDU 4970(杭电多校#9 1011题)Killing Monsters(瞎搞)
题目地址:HDU 4970 先进行预处理.在每一个炮塔的火力范围边界标记一个点. 然后对每一个点的伤害值扫一遍就能算出来. 然后在算出每一个点到终点的总伤害值,并保存下来,也是扫一遍就可以. 最后在询 ...
- hdu6373 Pinball 杭电第六场 物理知识
Pinball Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total S ...
随机推荐
- (转)sudo配置文件/etc/sudoers详解及实战用法
sudo配置文件/etc/sudoers详解及实战用法 原文:http://blog.csdn.net/field_yang/article/details/51547804 一.sudo执行命令的流 ...
- Shell分割字符得到数组
#!/bin/bash p=$(hadoop fs -ls /tgl/data |awk '{print $8}') #要将$a分割开,先存储旧的分隔符 OLD_IFS="$IFS" ...
- P4874 回形遍历 —模拟
思路: 写完后信心满满,结果超时. 我很不解,下了个数据结果——,z竟然是大于1e10的,跟题目给的不一样啊 原来如此,正解是一行一行的走的... 注意当到两边一样近时,应优先向下和右!!!!!! 这 ...
- AJPFX关于Java中的集合
ava API中所用的集合类,都是实现了Collection接口,他的一个类继承结构如下: Collection<--List<--Vector Collection<--List& ...
- ae(ArcEngine) java swing开发入门系列(2):ae的类型转换和Proxy类说明
做过C#版ae的都知道,操作同一个“对象”,用他的不同功能要转换到相应的接口,但java版有时不能直接做类型转换 例如下图在C#是可以的 但在java不行,这样转会报错,看IFeatureClass的 ...
- 【extjs6学习笔记】1.1 初始:创建项目
创建工作空间 sencha generate workspace /path/to/workspace 使用sencha创建应用 sencha -sdk /path/to/sdk generate a ...
- 用配置文件方式启动mongodb集群
- Ruby中访问控制符public,private,protected区别总结
重点关注private与protected public 默认即为public,全局都可以访问,这个不解释 private C++, “private” 意为 “private to this cla ...
- 3. Netbackup 7.6客户端的安装(windows/linux)
1 客户端的安装 1.1 Windows客户端安装 1.1.1 客户端hosts修改 windows xp/2003/vista/2008/7/8用户HOSTS文件是在“c:\windows\syst ...
- vue中的修饰符
Vue2.0学习笔记:Vue事件修饰符的使用 事件处理 如果需要在内联语句处理器中访问原生DOM事件.可以使用特殊变量$event,把它传入到methods中的方法中. 在Vue中,事件修饰符处理 ...