山东省第四届ACM大学生程序设计竞赛 Rescue The Princess-优快云博客

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在一场迷宫逃脱挑战中，一位王子必须利用几何知识找到通往公主所在位置的出口，通过解决数学问题来帮助自己和公主重获自由。

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Rescue The Princess

Time Limit: 1000MS Memory limit: 65536K

题目描述

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?

输入

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.

输出

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.

示例输入

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

示例输出

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

提示

来源

2013年山东省第四届ACM大学生程序设计竞赛

示例程序

tips：

atan 和 atan2 区别：

1：参数的填写方式不同；

2：atan2 的优点在于如果 x2-x1等于0 依然可以计算，但是atan函数就会导致程序出错；

ACcode：

#include <map>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define maxn 100
#define PI  acos(-1.0)
using namespace std;
int main(){
    double x1,x2,x3,y1,y2,y3,du,l;
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
        du=atan2((y2-y1),(x2-x1));
        l=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
        x3=x1+l*cos(du+PI/3.0);
        y3=y1+l*sin(du+PI/3.0);
        printf("(%.2lf,%.2lf)\n",x3,y3);
    }
    return 0;
}