Arthur's Table
题号:NC226652
时间限制:C/C++/Rust/Pascal 1秒,其他语言2秒
空间限制:C/C++/Rust/Pascal 256 M,其他语言512 M
Special Judge, 64bit IO Format: %lld

题目描述

King Arthur is well known for holding that all knights are equal and for his round table. There is a central serving platter (did Merlin invent the lazy-susan?) and each knight has a circular trencher to eat from:
Unfortunately, Merlin accidentally called up a pre-shade of George Orwell and Arthur started muttering something like:
                 … but some knights are more equal than others… 
So he asked Merlin to move the center platter away from him a bit so he and his closaest aides could have a bit more room:
Merlin complained that all the trenchers and the serving platter would have to be remade. Arthur, like a good leader said “Deal with it.”

Write a program to help Merlin make the new trenchers and serving platter. To describe the layout we will use a coodinate system with origin at the center of the table, with the x axis pointing to Arthur’s seat and the y axis perpendicular: 

输入描述:

Input consists of a single line containing the diameter, D, (8 <= D <= 30) of the table (in kingly feet),  the number, N, (7<= N <= 40) of knights to be seated, and the offset, O, 0.1 <= O < radius of an  original trencher) from the center of the table to the center of the central serving platter (as a decimal  number of kingly feet). 

输出描述:

The output consists five lines. The first output line gives the radius of the central platter (in kingly  feet) to 3 decimal places with center at (-offset, 0). (offset is O from the input.) The next four lines consist of three space separated floating point values to three decimal places.  These values represent the x coordinate of the center of a trencher, the y coordinate of the center of  the trencher and the radius of the trencher (all in kingly feet) of Arthur’s trencher and the next three  trenchers counter-clockwise from Arthur’s (one set of values for each line of output).
示例1

输入

复制
10 11 0.5

输出

复制
2.761
3.631 0.000 1.369
2.703 2.513 1.310
0.572 3.785 1.172
-1.625 3.621 1.031

备注:

浮点误差精度为 3 位(1e-3),fabs(stdvalue - yourvalue) / max(1.0, fabs(stdvalue)) < 1e-3均正确