Rescue The Princess（数学）

Problem A:Rescue The Princess

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
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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 equilateraltriangle 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.

Sample 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

Sample Output

(-50.00,86.60)

(-86.60,50.00)

(-36.60,136.60)

(1.00,1.00)

       题意：

       给出 T，代表有 T（1 ~ 100） 个 Test。每个 Test 给出 x1，y1 ，x2，y2，（绝对值 <= 1000）代表等边三角形的其中两个点，要求输出这个等边三角形第三个点的坐标，逆时针方向。

       思路：

       数学。求出两个点的终点坐标 （a，b）故可以得出方程，设给出两点构成的直线斜率为 k，与该指向垂直并经过（a，b）点的直线为 k1，等边三角形边长为 d：

       r = sqrt（3） * d / 2；

       k = （y1 - y2） / （x1 - x2）；

       k * k1 = -1；

       （x3 - a）^ 2 + （y3 - b）^ 2 = r ^ 2；

      （y3 - b）/ （x3 - a） = k1；

       由这些式子即可得出 x3，y3。并且对 x1 == x2 和 y1 == y2 进行另外讨论。

       AC：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
double dis(double x1, double y1, double x2, double y2) {
double xx = (x1 - x2) * (x1 - x2);
double yy = (y1 - y2) * (y1 - y2);
return sqrt(3) * sqrt(xx + yy) / 2;
}
int main () {
int t;
scanf("%d", &t);
while (t--) {
double x1, y1, x2, y2;
double d;
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
d = dis(x1, y1, x2, y2);
if (x1 == x2)
printf("(%.2f,%.2f)n", x1 + (y1 > y2 ? d : -d), (y1 + y2) / 2);
else if (y1 == y2)
printf("(%.2f,%.2f)n", (x1 + x2) / 2, y1 + (x1 > x2 ? -d : d));
else {
double a = (x1 + x2) / 2, b = (y1 + y2) / 2;
double k = (y1 - y2) / (x1 - x2);
double xa, xb, ya, yb;
xa = a + sqrt((k * k * d * d) / (1 + k * k));
xb = a - sqrt((k * k * d * d) / (1 + k * k));
ya = b - (xa - a) / k;
yb = b - (xb - a) / k;
y1 < y2 ? printf("(%.2f,%.2f)n", xb, yb) :
printf("(%.2f,%.2f)n", xa, ya) ;
}
}
return 0;
}