Rotate

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 573 Accepted Submission(s): 275
Special Judge

Problem Description

Noting is more interesting than rotation!

Your little sister likes to rotate things. To put it easier to analyze, your sister makes n rotations. In the i-th time, she makes everything in the plane rotate counter-clockwisely around a point ai by a radian of pi.

Now she promises that the total effect of her rotations is a single rotation around a point A by radian P (this means the sum of pi is not a multiplier of 2π).

Of course, you should be able to figure out what is A and P :).

Input

The first line contains an integer T, denoting the number of the test cases.

For each test case, the first line contains an integer n denoting the number of the rotations. Then n lines follows, each containing 3 real numbers x, y and p, which means rotating around point (x, y) counter-clockwisely by a radian of p.

We promise that the sum of all p's is differed at least 0.1 from the nearest multiplier of 2π.

T<=100. 1<=n<=10. 0<=x, y<=100. 0<=p<=2π.

Output

For each test case, print 3 real numbers x, y, p, indicating that the overall rotation is around (x, y) counter-clockwisely by a radian of p. Note that you should print p where 0<=p<2π.

Your answer will be considered correct if and only if for x, y and p, the absolute error is no larger than 1e-5.

Sample Input

Sample Output

1.8088715944 0.1911284056 3.0000000000

Source

2014 ACM/ICPC Asia Regional Anshan Online

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題意：給你n個點和角度，平面以點爲中心，旋轉一個給定的角度。然後把它們合併成由一個點和角度旋轉而來，輸出這個點和角度。

思路

1. 由平面旋轉知識：http://blog.csdn.net/wcyoot/article/details/33310329

所以構造矩陣，來求最終旋轉角度，及2個點座標。平面上任意一個點（x，y）以點（x0，y0）爲中心旋轉p角度，得到一個新點（X，Y）。

所以新點公式：

X=(x-x0)*cos(p)-(y-y0)*sin(p)+x0;

Y=(x-x0)*sin(p)+(y-y0)*cos(p)+y0;

兩個未知數，兩個式子，很容易求出來。

2. 還有一種思路是任取平面上兩個不同的點A1，A2，所以多次旋轉之後得到兩個點B1，B2，A1和B1的中垂線與A2和B2的中垂線的交點就是所求座標，角度也就很容易的處理。

下面貼一下思路1的AC代碼：

#include<stdio.h>
#include<math.h>
#include<string.h>
#define pi acos(-1.0)
#define ee 0.000001

struct Matrix 
{
  double a[2][2];
}m1,m2,m3; //定義的三個矩陣； 

//矩陣乘法； 
Matrix multi(Matrix a,Matrix b)
{
   Matrix c;
   int i,j,k;
   for(i=0;i<2;i++)
   for(j=0;j<2;j++)
   {
      c.a[i][j]=0;
      for(k=0;k<2;k++)
      c.a[i][j]+=a.a[i][k]*b.a[k][j];
   }
   return c;
}

int main()
{
   int T,i,n;
   double x,y,p,t,k1,k2;
   double xx,yy;
   scanf("%d",&T);
   while(T--)
   {
      scanf("%d",&n);
      m3.a[0][0]=m3.a[1][1]=1;
      m3.a[0][1]=m3.a[1][0]=0;
      m1.a[0][0]=m1.a[1][0]=m1.a[0][1]=m1.a[1][1]=0;
      for(i=0;i<n;i++)
      {
         scanf("%lf%lf%lf",&x,&y,&p);
         t=2*pi;
         while(p>=t)
         p-=t;
         m1.a[0][0]-=x;
         m1.a[0][1]-=y;
         m1.a[1][0]=m1.a[1][0]=0;
         m2.a[0][0]=m2.a[1][1]=cos(p);
         m2.a[0][1]=sin(p);
         m2.a[1][0]=-sin(p);
         m1=multi(m1,m2);
         m1.a[0][0]+=x;
         m1.a[0][1]+=y;
         m3=multi(m3,m2);
      }
      k1=m1.a[0][0];
      k2=m1.a[0][1];
      t=acos(m3.a[0][0]);
      if(sin(t)-m3.a[0][1]>ee) //防止精度損失； 
      t=2*pi-t;
      
      //解二元一次方程； 
      yy=(k1*sin(t)+k2*(1-cos(t)))/(2-2*cos(t));
      xx=(k1-yy*sin(t))/(1-cos(t));
      printf("%.6lf %.6lf %.6lf\n",xx,yy,t);
   }
   return 0;
}