Rescue The Princess（山東省第四屆ACM大學生程序設計競賽 ）

Problem Description

    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?

Input

    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.

Output

    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.

Example Input

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

Example Output

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

題意：等邊三角形，按逆時針順序給出a，b兩個點，讓你求c點；

思路：純數學問題；

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <math.h>
#include<cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <stack>
#define LL long long
#define INF 0x7fffffff
#define MAX 200010
#define PI 3.1415926535897932
using namespace std;

int main()
{
    int T;
    scanf("%d",&T);
    double x1,y1,x2,y2;
    double x,y;
    while(T--)
    {
        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
        if(abs(x1-x2) ==0)
        {
            if(y1<y2)
                x = x1- (sqrt(3.0) * (abs(y2-y1)/2));
            if(y1>y2)
                x = x1+ (sqrt(3.0) * (abs(y2-y1)/2));
            y = (y1+y2)/2;
            printf("(%.2lf,%.2lf)\n",x,y);
        }
        else
        {
            double L = sqrt( (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));
            double r1 = atan2( (y2-y1),(x2-x1));//包含正負號
            double r2 = PI/3 + r1;
            x = cos(r2)*L + x1;
            y = sin(r2)*L + y1;
            printf("(%.2lf,%.2lf)\n",x,y);
        }
    }
    return 0;
}