CANTON  Count on Cantor
One of the famous proofs of modern mathematics is Georg Cantor's demonstration that the set of rational numbers is enumerable. The proof works by using an explicit enumeration of rational numbers as shown in the diagram below.
1/1 1/2 1/3 1/4 1/5 ... 2/1 2/2 2/3 2/4 3/1 3/2 3/3 4/1 4/2 5/1
In the above diagram, the first term is 1/1, the second term is 1/2, the third term is 2/1, the fourth term is 3/1, the fifth term is 2/2, and so on.
Input
The input starts with a line containing a single integer t <= 20, the number of test cases. t test cases follow.
Then, it contains a single number per line.
Output
You are to write a program that will read a list of numbers in the range from 1 to 10^7 and will print for each number the corresponding term in Cantor's enumeration as given below.
Example
Input: 3 3 14 7 Output: TERM 3 IS 2/1 TERM 14 IS 2/4 TERM 7 IS 1/4
hide comments
manish_thakur:
20191212 12:28:09
position(x,y) = (1/2)(x+y)(x+y+1) + y


gak:
20191028 10:53:57
Observe the pattern by summing numerator and denominator, after that it's a cakewalk. 

saraswat000:
20190827 21:08:22
AC in one go! 

mriow:
20190820 09:02:27
Solution can be achieved in O(1) for every Cantor's Term.


jumaruba:
20190801 19:28:34
Hint:


maverick080800:
20190625 14:16:34
chaman chutiyo its so easy or sab ac in one go kr rhe ho . hurrrrr noobs 

urjajn123:
20190514 14:22:27
Last edit: 20190515 11:59:29 

aj_254:
20190511 17:27:49
just obsever the pattern and apply stl binary search .. easy and solvable in python.. 

ashimk:
20190318 20:26:34
Submitted after making a table upto the maximum value of number i.e 10^7 but got R.E(which was obvious though) . Then generated the number for every n.


newbie_127:
20190312 17:49:42
Use std :: lower_bound( ) . AC in 1 go :) 
Added by:  ThanhVy Hua 
Date:  20050227 
Time limit:  5s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: NODEJS PERL6 VB.NET 
Resource:  ACM South Eastern European Region 2004 