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).

#include <iostream>

#include <cmath>

#include <cstdio>

using namespace std;

int main()

{

    int T;

    double px,tx,t,ty,sum,a;

    cin>>T;

    while(T--)

    {

        cin>>px>>tx>>a;

        t=(tan(a)*px)/(tx*tx-2.0*tx*px);

        ty=t*tx*tx+tx*tan(a);

        sum=(0.5*(px-tx)*ty)+(/3.0*t*tx*tx*tx+0.5*tan(a)*tx*tx);

        printf("%.3lf\n",sum);

    }

    return ;

}

//题意：求由实线O-Tappingposition-Pig-O所围成图形的面积 s.

#include<stdio.h>

#include<math.h>

int main()

{

    int n;

    scanf("%d",&n);

    while(n--)

    {

        int t,p;

        double a,t1,t2;

        scanf("%d%d%lf",&p,&t,&a);

        t1=p*t*(*p-*t);

        t2=*(*p-t);

        printf("%.3lf\n",t1/t2*tan(a));

    }

    return ;

}

/*由题意可设抛物线方程为f(x)=a*x^2+b*x ，Tap点的纵坐标为 y,

由O-Tappingposition-Tx-O所围成图形的面积为 s1，

由Tx-Tappingposition-pig-Tx所围成图形的面积为s2.

f'(x)=2*a*x+b

s=s1+s2  ...... (1)

s2=1/2*(px-tx)*y  ...... (2)

s1=1/3*a*tx^3+1/2*b*tx^2  ...... (3)

f'(0)=tan(a) => b=tan(a)  ...... (4)

f(tx)=y => a*tx^2+b*tx=y  ...... (5)

f'(tx)=-y/(px-tx) => 2*a*tx+b=-y/(px-tx)  ...... (6)

联立(1)(2)(3)(4)(5)(6)解得：s=[px*tx*(3*px-2*tx)]/[6*(2*px-tx)]*tan(a)*/