CANTON - Count on Cantor

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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

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.


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.


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.



TERM 3 IS 2/1
TERM 14 IS 2/4
TERM 7 IS 1/4

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urjajn123: 2019-05-14 14:22:27

Last edit: 2019-05-15 11:59:29
aj_254: 2019-05-11 17:27:49

just obsever the pattern and apply stl binary search .. easy and solvable in python..

ashimk: 2019-03-18 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.
Just observe the pattern and BOOM!! A.C.

newbie_127: 2019-03-12 17:49:42

Use std :: lower_bound( ) . AC in 1 go :)

anant6025: 2019-02-05 22:02:11

***Hints Below***
Hint 1: [edited]

Hint 2: [edited]
Hint 3: [edited by Francky - read the notes below, point 3]

Easy Problem once you figure out the pattern.
AC in one go!

Last edit: 2019-02-06 07:16:41
cypher33: 2018-12-27 06:15:56

@abhinav__ your comment made my day lmao

adipat: 2018-10-12 19:55:37

This was how I solved it. First identify the diagonal pattern. A diagonal k has k terms and the total terms upto that diagonal will be (1+2+...+k). Using this property, locate the diagonal in which n will lie. From there the exact row and column of the nth term can be found by iterating through the diagonal up or down depending on whether it's an odd or even diagonal.

rajat_enzyme: 2018-08-20 19:58:32

Last edit: 2018-08-20 20:01:19
quannguyenlhp: 2018-08-18 06:27:54

AC in 1000000th go, 5 sec

Last edit: 2018-08-18 06:28:15
harsh771: 2018-07-08 19:02:48

Easy to solve once you observe the pattern!

Added by:Thanh-Vy Hua
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