Description
Several days ago, a beast caught a beautiful princess and the princess was put in *. 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.
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).
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：
给你等边三角形的两个点A和B，求第三个点C的坐标；
且ABC是逆时针的；
思路：
因为要求ABC是逆时针的，所以可以直接用B绕A逆时针旋转60°；
这里有个通用的公式，证明稍微复杂，可以加到模板里以备不时之需：
点（x1，y1）绕点（x2，y2）逆时针旋转a角度后新的坐标（X，Y）为:
X=(x1-x2)*cos(a)-(y1-y2)*sin(a)+x2;
Y=(x1-x2)*sin(a)+(y1-y2)*cos(a)+y2;
如果直接按照题意的等边三角形的情况去画图推导也可以推导出来，不过这个公式比较普适。
代码：
#include <bits/stdc++.h>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <queue>
#include <stack>
#include <map>
#include <set> #define IO ios::sync_with_stdio(false);\
cin.tie();\
cout.tie(); typedef long long LL;
const long long INF = 0x3f3f3f3f;
const long long mod = 1e9+;
const double PI = acos(-1.0);
const int maxn = ;
const char week[][]= {"Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"};
const char month[][]= {"Janurary","February","March","April","May","June","July",
"August","September","October","November","December"
};
const int daym[][] = {{, , , , , , , , , , , , },
{, , , , , , , , , , , , }
};
const int dir4[][] = {{, }, {, }, {-, }, {, -}};
const int dir8[][] = {{, }, {, }, {-, }, {, -}, {, }, {-, -}, {, -}, {-, }}; int main() {
int t;
scanf("%d", &t);
while(t--){
double x1,x2,x3,y1,y2,y3;
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
double dx=x2-x1,dy=y2-y1;
x3=dx/-dy*sqrt(3.0)/+x1;
y3=dy/+dx*sqrt(3.0)/+y1;
printf("(%.2lf,%.2lf)\n",x3,y3);
}
return ;
}