山东省第四届ACM程序设计竞赛A题：Rescue The Princess(向量旋转)

A - Problem A:Rescue The Princess

Time Limit:1000MS     Memory Limit:131072KB     64bit IO Format:%lld & %llu

Submit Status

 use MathJax to parse formulas

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.

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)

Hint

这个题想赢求坐标不是很方便，最好还是用向量旋转的方法

我们令a（x1,y1）,b(x2,y2),角度为A

则

公式：

           x3 = （x2-x1）*cosA-(y2-y1)*sinA+x1;

            y3 =     (x2-x1)*sinA - (y2-y1)*cosA + y1;

代码如下

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 110
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
typedef long long ll;
int gcd(int a,int b){return b?gcd(b,a%b):a;}

int main()
{
    double r,x1,x2,y1,y2;
    double x3,y3;
    int n;
    cin >> n;
    while(n--){
        cin >> x1>>y1>>x2>>y2;
        x3 = (x2-x1)/2-(y2-y1)/2*sqrt(3)+x1;
        y3 = (y2-y1)/2+(x2-x1)/2*sqrt(3)+y1;
        printf("(%.2f,%.2f)\n",x3,y3);
 
    }
    return 0;
}