zoukankan      html  css  js  c++  java
  • HDU 6373 Pinball

    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*(1+b*b*1.0/(a*a)));
            double TR=sqrt(2*l/(g*SIN)),T=sqrt(2*h/g);
            int cnt=int(TR/T),ans=(cnt+1)>>1;
            cout<<ans<<endl;
        }
    }
    int main() {
        init();
        solve();
        return 0;
    }
    View Code
  • 相关阅读:
    javaSE基础知识点
    java自定义注解
    java自定义线程池
    java写投票脚本自动化初探
    java线程安全初窥探
    锁的深入理解
    java守护线程与非守护线程
    java设计模式初探
    java内存模型初窥探
    uniapp中组件之间跳转遇到的问题
  • 原文地址:https://www.cnblogs.com/Rhythm-/p/9445257.html
Copyright © 2011-2022 走看看