山東省第四屆 A Rescue The Princess

題目描述

    Several days ago, a beast caught a beautiful princess and the princess was put in prison. To rescue the princess, a prince who wanted to marry the princess set out immediately. Yet, the beast set a maze. Only if the prince find out the maze’s exit can he save the princess.

    Now, here comes the problem. The maze is a dimensional plane. The beast is smart, and he hidden the princess snugly. He marked two coordinates of an equilateral triangle in the maze. The two marked coordinates are A(x₁,y₁) and B(x₂,y₂). The third coordinate C(x₃,y₃) is the maze’s exit. If the prince can find out the exit, he can save the princess. After the prince comes into the maze, he finds out the A(x₁,y₁) and B(x₂,y₂), but he doesn’t know where the C(x₃,y₃) is. The prince need your help. Can you calculate the C(x₃,y₃) and tell him?

輸入

    The first line is an integer T(1 <= T <= 100) which is the number of test cases. T test cases follow. Each test case contains two coordinates A(x₁,y₁) and B(x₂,y₂), described by four floating-point numbers x₁, y₁, x₂, y₂ ( |x₁|, |y₁|, |x₂|, |y₂| <= 1000.0).
    Please notice that A(x₁,y₁) and B(x₂,y₂) and C(x₃,y₃) are in an anticlockwise direction from the equilateral triangle. And coordinates A(x₁,y₁) and B(x₂,y₂) are given by anticlockwise.

輸出

    For each test case, you should output the coordinate of C(x₃,y₃), the result should be rounded to 2 decimal places in a line.

示例輸入

4
-100.00 0.00 0.00 0.00
0.00 0.00 0.00 100.00
0.00 0.00 100.00 100.00
1.00 0.00 1.866 0.50

示例輸出

(-50.00,86.60)
(-86.60,50.00)
(-36.60,136.60)
(1.00,1.00)

#include <stdio.h>
#include <math.h>
const double pi = acos(-1.0);
int main()
{
    int t;
    double x1,x2,x3,y1,y2,y3,l,at;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
        at = atan2(y2-y1,x2-x1);
        l = sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
        x3 = x1+l*cos(at+pi/3.0);
        y3 = y1+l*sin(at+pi/3.0);
        printf("(%.2lf,%.2lf)\n",x3,y3);
    }
    return 0;
}

漸漸體會到了數學的可愛and可怕之處了。。。

也知道爲什麼大部分大神都是深井冰狀態。。。

知識：

C語言中的atan和atan2

在C語言的math.h或C++中的cmath中有兩個求反正切的函數atan(double x)與atan2(double y,double x)  他們返回的值是弧度 要轉化爲角度再自己處理下。

前者接受的是一個正切值（直線的斜率）得到夾角，但是由於正切的規律性本可以有兩個角度的但它卻只返回一個，因爲atan的值域是從-90~90 也就是它只處理一四象限，所以一般不用它。

第二個atan2(double y,double x) 其中y代表已知點的Y座標 同理x ,返回值是此點與遠點連線與x軸正方向的夾角，這樣它就可以處理四個象限的任意情況了，它的值域相應的也就是-180~180了

例如：

例1：斜率是1的直線的夾角

cout<<atan(1.0)*180/PI;//45°

cout<<atan2(1.0,1.0)*180/PI;//45° 第一象限

cout<<atan2(-1.0,-1.0)*180/PI;//-135°第三象限

後兩個斜率都是1 但是atan只能求出一個45°

例2：斜率是-1的直線的角度

cout<<atan(-1.0)*180/PI;//-45°

cout<<atan2(-1.0,1.0)*180/PI;//-45° y爲負 在第四象限

cout<<atan2(1.0,-1.0)*180/PI;//135° x爲負 在第二象限

 

常用的不是求過原點的直線的夾角 往往是求一個線段的夾角 這對於atan2就更是如魚得水了

例如求A(1.0,1.0) B(3.0,3.0)這個線段AB與x軸正方向的夾角

用atan2表示爲 atan2(y2-y1,x2-x1) 即 atan2(3.0-1.0,3.0-1.0)

它的原理就相當於把A點平移到原點B點相應變成B'（x2-x1,y2-y1）點 這樣就又回到先前了

例三：

A(0.0,5.0) B(5.0,10.0)

線段AB的夾角爲

cout<<atan2(5.0,5.0)*180/PI;//45°

^_^