HDU 4998 Rotate

Rotate

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1116    Accepted Submission(s): 503
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

130 0 11 1 12 2 1

 

Sample Output

1.8088715944 0.1911284056 3.0000000000

 

Source

2014 ACM/ICPC Asia Regional Anshan Online

 

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计算几何，有一个点，绕着另一个点旋转了p角度。并且旋转了n次。问最初这个点等价于绕哪个点旋转n角度和那个等价。

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<string>#include<algorithm>using namespace std;#define N 100050typedef long long ll;const int MOD = 1e9+7;#define PI acos(-1)int  main(){    double start_x, start_y;    double final_x, final_y;    double x, y, p;    double  final_jiao, xx, yy;    int t;    scanf("%d",&t);    while(t--)    {        int n;        scanf("%d",&n);        start_x=start_y=xx=yy=1;  //随便设立        final_jiao=0;        while(n--)        {            scanf("%lf %lf %lf",&x,&y,&p);            final_jiao+=p;            if(final_jiao>=2*PI)                final_jiao-=2*PI;            final_x=(xx-x)*cos(p)-(yy-y)*sin(p)+x;    // x0= (x - xx)*cos(a) - (y - yy)*sin(a) + xx ;  设x,y绕着点xx,yy转了p角度。则x0,y0为新的点。            final_y=(xx-x)*sin(p)+(yy-y)*cos(p)+y;   //  y0= (x - xx)*sin(a) + (y - yy)*cos(a) + yy ;            xx=final_x;            yy=final_y;        }        // 换下公式，相当于在一开始的点然后找一个点旋转使得最后的点为fianl_x,final_y  //则可以推出        x=((final_x-start_x*cos(final_jiao)+start_y*sin(final_jiao))*(1-cos(final_jiao))-(final_y-start_x*sin(final_jiao)-start_y*cos(final_jiao))*sin(final_jiao))/(2-2*cos(final_jiao));        y=((final_x-start_x*cos(final_jiao)+start_y*sin(final_jiao))*(1-cos(final_jiao))-(1-cos(final_jiao))*(1-cos(final_jiao))*x)/((1-cos(final_jiao))*sin(final_jiao));        printf("%.12lf %.12lf %.12lf\n",x,y,final_jiao);    }    return 0;}