[ACM] sdut 2877 angry_birds_again_and_again (简单数学积分）_8572. 木棒裁切 time limit: 2000ms memory limit: 1024k -优快云博客

本文探讨如何计算游戏《愤怒的小鸟》中黄色小鸟Chuck路径与地面围成的面积。通过数学公式推导，利用抛物线飞行轨迹及其切线特性，结合给定角度、坐标等参数，给出具体的计算方法及示例。

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angry_birds_again_and_again

Time Limit: 2000ms Memory limit: 65536K 有疑问？点这里^_^

题目描述

The problems called "Angry Birds" and "Angry Birds Again and Again" has been solved by many teams in the series of contest in 2011 Multi-University Training Contest.

This time we focus on the yellow bird called Chuck. Chuck can pick up speed and distance when tapped.

You can assume that before tapped, Chuck flies along the parabola. When tapped, it changes to fly along the tangent line. The Chuck starts at the coordinates (0, 0). Now you are given the coordinates of the pig (Px, 0), the x-coordinate of the tapping position (Tx) and the initial flying angle of Chuck (α).

∠AOx = α

Please calculate the area surrounded by Chuck’s path and the ground.(The area surrounded by the solid line O-Tapping position-Pig-O)

输入

The first line contains only one integer T (T is about 1000) indicates the number of test cases. For each case there are two integers, px tx, and a float number α.（0 < Tx ≤ Px ≤ 1000, 0 < α <

） .

输出

One line for each case specifying the distance rounded to three digits.

示例输入

1
2 1 1.0

示例输出

0.692

提示

来源

2014年山东省第五届ACM大学生程序设计竞赛

解题思路：

设方程 y=ax^2+bx+c ，图中曲线经过原点，所以c=0.

对方程求导 y'=2ax+b , y'代表斜率，那么原点(0,0)这一点，代人y'=b,即该点的斜率，根据题意b=tan( α）

如图：在题目中x=tx这一点时，容易混，记tx为t，图中曲线x=t这一点，该点的斜率为 2at+b . 注意斜率是负的

三角形竖着的直角边除以横着的直角边（p-t)的值的相反数即为斜率 2at+b

竖着的直角边值为 at^2+bt (将t带入原方程）,横着的直角边为p-t，则有式子

2at+b= - ( at^2+bt)/(p-t)

解出a，这样方程中a,b的值都有了。

那么题目所求的面积即为曲线覆盖面积从 0到t积分积分函数为（ax^2+bx) ，再加上三角形的面积 0.5*(p-t)*(at^2+bt)

代码：

#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;

double px,tx,jiao;
double a,b;

int main()
{
    int t;cin>>t;
    while(t--)
    {
        cin>>px>>tx>>jiao;
        b=tan(jiao);
        double m=px-tx;
        a=(-b*tx-b*m)/(2*tx*m+tx*tx);
        double ans;
        ans=(1/3.0)*a*tx*tx*tx+0.5*b*tx*tx+0.5*(px-tx)*(a*tx*tx+b*tx);
        cout<<setiosflags(ios::fixed)<<setprecision(3)<<ans<<endl;
    }
    return 0;
}