CUBEFR  Cube Free Numbers
A cube free number is a number who’s none of the divisor is a cube number (A cube number is a cube of a integer like 8 (2 * 2 * 2) , 27 (3 * 3 * 3) ). So cube free numbers are 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18 etc (we will consider 1 as cube free). 8, 16, 24, 27, 32 etc are not cube free number. So the position of 1 among the cube free numbers is 1, position of 2 is 2, 3 is 3 and position of 10 is 9. Given a positive number you have to say if its a cube free number and if yes then tell its position among cube free numbers.
Input
First line of the test case will be the number of test case T (1 <= T <= 100000) . Then T lines follows. On each line you will find a integer number n (1 <= n <= 1000000).
Output
For each input line, print a line containing “Case I: ”, where I is the test case number. Then if it is not a cube free number then print “Not Cube Free”. Otherwise print its position among the cube free numbers.
Example
Sample Input: 10 1 2 3 4 5 6 7 8 9 10 Sample Output: Case 1: 1 Case 2: 2 Case 3: 3 Case 4: 4 Case 5: 5 Case 6: 6 Case 7: 7 Case 8: Not Cube Free Case 9: 8 Case 10: 9
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_:
20130521 13:35:18
i dont know why i am getting wrong ans.. checked many cases..


Santiago Palacio:
20130202 18:10:58
Jmm... scanf must be used for this one. 

saket diwakar:
20130117 21:38:52
nice one...:) 

J.A.R.V.I.S.:
20121210 07:06:27
@Adhityaa


Akash Bansal:
20121026 17:41:23
what is the answer for n=99998


eliminator:
20120929 14:22:19
linear search gives TLE...


Adhityaa:
20120927 08:12:47
How is 1 cube free (1 = 1*1*1) ? 

vipul gupta:
20120925 05:54:40
can't tell why i am getting WA. please check code no. 7718647 for this problem... i made many test cases and all have given correct answer even the extreme one. please tell.


Rocker3011:
20120618 03:40:40
a smart precompute is enough for this problem :D 
Added by:  Muhammad Ridowan 
Date:  20110614 
Time limit:  0.100s1s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: ASM64 
Resource:  Own. For alternate thanks Sayef Azad Sakin 