Rotate

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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

1
3
0 0 1
1 1 1
2 2 1

Sample Output

1.8088715944 0.1911284056 3.0000000000

Source

2014 ACM/ICPC Asia Regional Anshan Online

题目大意：给出n个点及旋转角度p（P是弧度制），问二维平面的点分别绕着这n个点旋转对应的角度后，相当于整个二维平面绕那个点旋转多少弧度，误差不超过10^(-5)。

解题思路：

对于P（x，y）点绕定点A（x0，y0）旋转角度p，得到新的点 P’（X，Y），则有下列公式：

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

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

我们不妨找一个不在数据范围内的点P（-1，-20），对该点做n次旋转变换，得到P‘（X，Y），而一次的旋转角SP就是n个p的加和。

则对于P、P’和SP，我们可以按照上面的旋转公式列一个二元一次方程组，解方程组即可。

代码如下：

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <fstream>
#include <limits.h>
#include <ctime>
#define debug "output for debug\n"
#define pi (acos(-1.0))
#define eps (1e-6)
#define inf (1<<28)
#define sqr(x) (x) * (x)
#define mod 1000000007
using namespace std;
typedef long long ll;
typedef unsigned long long ULL;
struct point
{
    double x;
    double y;
    double p;
};
point P[15];
//旋转公式
point XUAN_ZHUAN(point a,point b,double p)
{
    point c;
    c.x=(b.x-a.x)*cos(p)-(b.y-a.y)*sin(p)+a.x;
    c.y=(b.x-a.x)*sin(p)+(b.y-a.y)*cos(p)+a.y;
    return c;
}
int main()
{
    int i,n,t;
    point p0,p1;
    double sp;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        //寻找的P（-1，-20）点
        p0.x=p1.x=-1.0;p0.y=p1.y=-20.0;
        sp=0.0;
        //n次旋转
        for(i=0;i<n;i++)
        {
            scanf("%lf%lf%lf",&P[i].x,&P[i].y,&P[i].p);
            p1=XUAN_ZHUAN(P[i],p1,P[i].p);
            sp+=P[i].p;
            while(sp>=2*pi)
                sp=sp-(2*pi);
        }
        //二元一次方程组的解
        double y=(p1.x*sin(sp)+p1.y*(1-cos(sp))-(p0.x*cos(sp)-p0.y*sin(sp))*sin(sp)-(p0.x*sin(sp)+p0.y*cos(sp))*(1-cos(sp)))/(2-2*cos(sp));
        double x=(p1.x*(1-cos(sp))-p1.y*sin(sp)-(p0.x*cos(sp)-p0.y*sin(sp))*(1-cos(sp))+(p0.x*sin(sp)+p0.y*cos(sp))*sin(sp))/(2-2*cos(sp));
        printf("%.6lf %.6lf %.6lf\n",x,y,sp);
    }
    return 0;
}

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