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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 107 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
| Added by: | Thanh-Vy Hua |
| Date: | 2005-02-27 |
| 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 |
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2019-08-20 09:02:27
Solution can be achieved in O(1) for every Cantor's Term. Hint: Use the formula for the Sum of the First n Terms of an Arithmetic Sequence. |
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2019-08-01 19:28:34
Hint: Consider that it's a arithmetic progression second order. 0.0 second code |
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2019-06-25 14:16:34
chaman chutiyo its so easy or sab ac in one go kr rhe ho . hurrrrr noobs |
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2019-05-14 14:22:27
Last edit: 2019-05-15 11:59:29 |
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2019-05-11 17:27:49
just obsever the pattern and apply stl binary search .. easy and solvable in python.. |
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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. |
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2019-03-12 17:49:42
Use std :: lower_bound( ) . AC in 1 go :) |
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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 |
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2018-12-27 06:15:56
@abhinav__ your comment made my day lmao |
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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. |
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