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旋转

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 461    Accepted Submission(s): 318


Problem Description
求空间中一个点绕一穿过原点的轴线旋转一定角度后的坐标，沿着旋转轴往原点看旋转的角度为顺时针（Angles are measured clockwise when looking along the rotation axis toward the origin. ）。
 

Input
一个点的3个坐标 x,y,z,
轴线上除原点外的一点 xn,yn,zn
角度 angle（弧度）
 

Output
旋转以后的坐标 （小数点后保留3位）
 

Sample Input
2.0 3.0 6.0
5.0 5.0 5.0 
3.14 
100.0 200.0 300.0
500.0 600.0 700.0 
1.25
 

Sample Output
5.336 4.330 1.334
185.988 132.594 296.357
 

Source
2009 Multi-University Training Contest 10 - Host by NIT
 

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