连通图的稳定状态的计算用的数据结构,怎么采用C程序语言的编程算法的实现的过程

Problem Description
Rasen had lost in labyrinth for 20 years. In a normal day, he found a bright screen. There were 4 points labeled by ‘A’ , ‘B’ , ‘C’ , ‘D’, and rasen could drag these point. And two points ‘E’ , ‘F’ moved. Rasen found that ‘E’ is the middle point of ‘A’ and ‘B’, and ‘F’ is the middle point of ‘C’ and ‘D’. Near the screen there was a marble slab.There were a list of the distance of AB , BC , CD , DA and EF. Rasen also found that the distance of these edge of the points in screen showed at the same time he drop the points. He wanted to know what will happen if the distances in screen are same with the number in slab.

Input
The first line of input contains only one integer T(<=50000), the number of test cases.
Each case contains five float number, indicating the distance of AB , BC , CD , DA , EF.(0<=distance<=10000)

Output
For each test, first print a line “Case #i:”(without quotes),with i implying the case number, then output the coordinates of A,B,C,D four points. Answer will be considered as correct if the length got from your output (the spj will use double to get the point, and the distance from two points will calculate in the way of sqrt((x1-x2)^2 +(y1-y2)^2) ) and the length given is less than 10-4.
(It guarantees that there exists a solution. If there are many solutions, output any one of them.)

Sample Input
1
1.000000 1.000000 1.000000 1.000000 1.000000

Sample Output
Case #1:
0.000000 0.000000
1.000000 0.000000
1.000000 1.000000
0.000000 1.000000

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