MAXLN - THE MAX LINES

In this problem you will be given a half-circle. The half-circle’s radius is r. You can take any point A on the half-circle and draw 2 lines from the point to the two sides of the diameter(AB and AC). Let the sum of square of one line’s length and the other line’s length is s

Like in the figure s = AB2 + AC. And BC = 2r.

Now given r you have to find the maximum value of s. That is you have to find point A such that AB2 + AC is maximum.

Input

First line of the test case will be the number of test case T (1 <= T <= 1000). Then T lines follows. On each line you will find a integer number r (1 <= r <= 1000000); each representing the radius of the half-circle.

Output

For each input line, print a line containing "Case I: ", where I is the test case number and the maximum value of s. Print 2 digit after decimal (Errors should be less then .01).

Example

Sample Input:
1
1

Sample Output:
Case 1: 4.25

Added by:Muhammad Ridowan
Date:2011-03-28
Time limit:1s-1.679s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64
Resource:Own Problem(used for CSE,University of Dhaka, Newbies Contest)

hide comments
2017-01-15 22:07:59
Be careful of the output format.The 'C' in case is capital.Cost me 7 wa........
2017-01-01 18:40:08
My 50th !
2016-12-14 22:30:41
AC by hook or crook...
2016-12-14 14:31:07
mathematical concept required......
2016-12-04 09:28:14
care of format of output required.
2016-11-09 11:57:02 E Naveen Kumar
REMEMBER ans should be in format "Case I: ans"
It costed me 1 WA
2016-10-31 08:55:27
when you add that "fixed number" to the value of computed number,it gives WA but when you just print it it gives AC. Don't know why
2016-09-27 19:52:39
one WA for not printing "Case :".. took radius as double
2016-08-29 14:11:34
AC in one go :D. Comments do help. Take r as double.
2016-08-26 14:25:12
Case costed me 2 WA no diff needed just trigy :P
no double or float needed
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