http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2877
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
数学知识学得很不扎实，很大程度程度上是在应付考试，导致不会灵活运用，在高中这种题手到擒来，但现在感觉做的有点费事。
来源
2014年山东省第五届ACM大学生程序设计竞赛
#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 ;
}
简单数学题，大神的思路
//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)*/