SDUT 2603 Rescue The Princess

Rescue The Princess

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

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

 首先介绍下atan函数与atan2函数的一点区别

 

atan 和 atan2 都是求反正切函数，如：有两个点 point(x1,y1), 和 point(x2,y2);

那么这两个点形成的斜率的角度计算方法分别是：

float angle = atan( (y2-y1)/(x2-x1) );

或

float angle = atan2( y2-y1, x2-x1 );

 

 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；

 

结论： atan 和 atan2函数，建议用 atan2函数；

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>

using namespace std;

#define PI acos(-1.0)

int main(){

    //freopen("input.txt","r",stdin);

    int t;
    scanf("%d",&t);
    double a,b,a1,b1,a2,b2;
    while(t--){
        scanf("%lf%lf%lf%lf",&a1,&b1,&a2,&b2);
        double degree=atan2(b2-b1,a2-a1);
        double len=sqrt((b2-b1)*(b2-b1)+(a2-a1)*(a2-a1));
        a=a1+len*cos(degree+PI/3);
        b=b1+len*sin(degree+PI/3);
        printf("(%.2lf,%.2lf)\n",a,b);
    }
    return 0;
}