sdutoj 2603 Rescue The Princess

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2603

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

分析：

已知等边三角形的两个按逆时针给出的两个顶点，求第三个点。

官方代码：

#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);
        printf("%lf
",(y2-y1)/(x2-x1));
        printf("%lf
",at);
        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)
",x3,y3);
    }

    return 0;
}

AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
#define D double
#define PI acos(-1.0)
D x3,y3;
int d1, d2;
D Length(D x1,D y1,D x2,D y2 )
{
    return sqrt((y2-y1)*(y2-y1) + (x2-x1)*(x2-x1));
}
void Solve(D x1, D y1, D x2,D y2)
{
    D Radian = atan((y2 - y1)/(x2 - x1));
    D Angle = Radian * 180.0 / PI;
    if (x1 > x2)
        Angle += 180.0;
    Radian += PI/3.0;
    Angle += 60.0;
     if (Angle <= 90 || Angle >= 270)
         d1 = 1;
    else
        d1 = -1;
     if (Angle >= 0 && Angle <= 180)
         d2 = 1;
    else
        d2 = -1;
    D dis = Length(x1,y1,x2,y2);
    x3 = x1 + d1 * dis * fabs(cos(Radian));
    y3 = y1 + d2 * dis * fabs(sin(Radian));
}
int main()
{
    int T;
    D x1, y1, x2, y2;
    scanf ("%d", &T);
    while (T --)
    {
        scanf ("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
        Solve(x1, y1, x2, y2);
        printf ("(%.2lf,%.2lf)
", x3, y3);
    }

    return 0;
}

还可以推出公式的：

利用向量的旋转：http://www.cnblogs.com/jeff-wgc/p/4468038.html

AC代码：

#include<stdio.h>
#include<math.h>
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        double x1,x2,y1,y2,x3,y3,x,y;
        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
        x=x2-x1;
        y=y2-y1;
        x3=x/2-y*sqrt(3)/2+x1;
        y3=y/2+x*sqrt(3)/2+y1;
        printf("(%.2f,%.2f)
",x3,y3);
    }
    return 0;
}