Rescue The PrincessTime 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 anequilateraltrianglein 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 youcalculatethe 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 theequilateraltriangle. 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 0.000.00 0.00 0.00 100.000.00 0.00 100.00 100.001.00 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
题意：已知两点A,B，在A,B连线的逆时针方向上找一点C，使三点构成一个等边三角形。
思路：
1.求已知两点的斜率R。
2.求边长len。
3.求第三点 的坐标x3=x1+L*cos(α+60)；y3=y1+L*sin(α+60)【因为是在两个点的基础上再旋转60度，弱若以A点为原点，，所求点的坐标是后面的乘法不分，因为是要求绝对坐标，所以要加上A的坐标点】。
#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>using namespace std;typedef long long LL;const int inf=0x3f3f3f3f;const double pi= acos(-1.0);int main(){int n;double x1,y1,x2,y2,x3,y3;double R,len;scanf("%d",&n);while(n–){scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);R=atan2((y2-y1),(x2-x1));// 求反函数用atan2最好、len=sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));x3=x1+len*cos(R+pi/3);y3=y1+len*sin(R+pi/3);printf("(%.2lf,%.2lf)\n",x3,y3);}return 0;}
如果困难是堵砖墙，拍拍它说你还不够高。
