Pinball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 131    Accepted Submission(s): 55

Problem Description
There is a slope on the 2D plane. The lowest point of the slope is at the origin. There is a small ball falling down above the slope. Your task is to find how many times the ball has been bounced on the slope.

It's guarantee that the ball will not reach the slope or ground or Y-axis with a distance of less than 1 from the origin. And the ball is elastic collision without energy loss. Gravity acceleration $g = 9.8 m/s^2$.

 
Input
There are multiple test cases. The first line of input contains an integer T (1 $\le$ T $\le$ 100), indicating the number of test cases.

The first line of each test case contains four integers a, b, x, y (1 $\le$ a, b, -x, y $\le$ 100), indicate that the slope will pass through the point(-a, b), the initial position of the ball is (x, y).

 
Output
Output the answer.

It's guarantee that the answer will not exceed 50.

 
Sample Input
1
5 1 -5 3
 
Sample Output
2
 
Source
 
Recommend
chendu
 

Statistic | Submit | Discuss | Note

分析:高中物理题可还行,直接正交分解推个公式就好了。
#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <deque>
#include <map>
#define range(i,a,b) for(auto i=a;i<=b;++i)
#define LL long long
#define ULL unsigned long long
#define elif else if
#define itrange(i,a,b) for(auto i=a;i!=b;++i)
#define rerange(i,a,b) for(auto i=a;i>=b;--i)
#define fill(arr,tmp) memset(arr,tmp,sizeof(arr))
#define IOS ios::sync_with_stdio(false);cin.tie(0)
using namespace std;
int t,a,b,x,y;
const double g=9.8;
void init(){
scanf("%d",&t);
}
void solve(){
while(t--){
scanf("%d%d%d%d",&a,&b,&x,&y);
double SIN=b*1.0/sqrt(a*a+b*b);
double h=y+b*x*1.0/a,l=h*SIN+sqrt(x*x*(+b*b*1.0/(a*a)));
double TR=sqrt(*l/(g*SIN)),T=sqrt(*h/g);
int cnt=int(TR/T),ans=(cnt+)>>;
cout<<ans<<endl;
}
}
int main() {
init();
solve();
return ;
}

HDU 6373 Pinball的更多相关文章

  1. HDU 6373.Pinball -简单的计算几何+物理受力分析 (2018 Multi-University Training Contest 6 1012)

    6373.Pinball 物理受力分析题目. 画的有点丑,通过受力分析,先求出θ角,为arctan(b/a),就是atan(b/a),然后将重力加速度分解为垂直斜面的和平行斜面的,垂直斜面的记为a1, ...

  2. cdq分治(hdu 5618 Jam's problem again[陌上花开]、CQOI 2011 动态逆序对、hdu 4742 Pinball Game、hdu 4456 Crowd、[HEOI2016/TJOI2016]序列、[NOI2007]货币兑换 )

    hdu 5618 Jam's problem again #include <bits/stdc++.h> #define MAXN 100010 using namespace std; ...

  3. HDU 4247 Pinball Game 3D(cdq 分治+树状数组+动态规划)

    Pinball Game 3D Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) ...

  4. hdu 4742 Pinball Game 3D(三维LIS&amp;cdq分治&amp;BIT维护最值)

    Pinball Game 3D Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) ...

  5. hdu 4742 Pinball Game 3D 分治+树状数组

    离散化x然后用树状数组解决,排序y然后分治解决,z在分治的时候排序解决. 具体:先对y排序,solve(l,r)分成solve(l,mid),solve(mid+1,r), 然后因为是按照y排序,所以 ...

  6. HDU 6373(斜面上小球弹跳 运动分解)

    题意是给定两个点的位置,过原点引一条射线穿过第一个点,射线位置作为斜面位置,第二个点处令一小球自由落体,问小球能碰撞到斜面几次. 开始时想算出两次碰撞中小球沿斜面运动的距离,然后发现每一段距离会因为高 ...

  7. 2018HDU多校联赛第六场 6373 Pinball——水题&&物理题

    题意 给定一个斜面,从某处让一个小球作自由落体运动,求小球与斜面的碰撞次数(假设都为弹性碰撞). 分析 题图如下,x轴.y轴是虚拟的. 根据高中物理的套路,沿斜面方向分解重力加速度即可. #inclu ...

  8. 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 ...

  9. HDOJ 2111. Saving HDU 贪心 结构体排序

    Saving HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total ...

随机推荐

  1. PHP扩展--APC缓存安装与使用

    apc安装 wget http://pecl.php.net/get/APC-3.1.13.tgz tar zxvf APC-3.1.13.tgz cd APC-3.1.13 /usr/local/p ...

  2. 洛谷P2868 [USACO07DEC]观光奶牛 Sightseeing Cows

    题目描述 Farmer John has decided to reward his cows for their hard work by taking them on a tour of the ...

  3. PHP 练习3:租房子

    一.题目要求 二.题目做法 1.建立数据库 2.封装类文件 <?php class DBDA { public $fuwuqi="localhost"; //服务器地址 pu ...

  4. Python eval 函数说明

    eval(str [,globals [,locals ]]) -- 函数将字符串str当成有效Python表达式来求值,并返回计算结果. 例 :  eval('3+4')         ==> ...

  5. poj 2000 Gold Coins

    题目链接:http://poj.org/problem?id=2000 题目大意:求N天得到多少个金币,第一天得到1个,第二.三天得到2个,第四.五.六天得到3个....以此类推,得到第N天的金币数. ...

  6. Kaggle机器学习之模型集成(stacking)

    Stacking是用新的模型(次学习器)去学习怎么组合那些基学习器,它的思想源自于Stacked Generalization这篇论文.如果把Bagging看作是多个基分类器的线性组合,那么Stack ...

  7. Ubuntu安装pip

    首先打开终端 在终端输入:sudo apt-get install python-pip python-dev build-essential [+] 如果需要在Python3下安装pip,那么在py ...

  8. deepin安装metasploit

    [1]安装metasploit 1.curl https://raw.githubusercontent.com/rapid7/metasploit-omnibus/master/config/tem ...

  9. xrange和range的区别

    >>> print type(range(5)) <type 'list'> >>> print type(xrange(5)) <type 'x ...

  10. Java中基于HotSpot虚拟机的垃圾收集器

    名称 过程 优缺点 Serial 进行垃圾收集时,必须暂停其他所有的工作进程,直到它收集结束.是一个单线程收集器. Stop the world. 新生代收集器. 手工设置新生代的大小:-Xmn Ed ...