Mann

@Mann Hi!

Why does dual degree have less competition than normal one?

My question is short but you gotta give me a detailed explanation XD

@AjayMishra I have some doubts about a code

@cattt ?

I have this question : codechef.com/JUNE19B/problems/CHFING

@AjayMishra

We and Mann discussed exactly this problem here.

@cattt see this, he had the algorithm.

@AjayMishra pastebin.com/TWTs0s4v

That's quite long program for little work. What exactly is the problem by the way?

codechef.com/JUNE19B/problems/CHFING

Oh, You told me that you have problem with your program, I was expecting there me some error in the program. Isn't that the case?

@AjayMishra I think I am missing some case

Or maybe some code problem

Did you find that in the link(codechef...) the input/output do not conform with the goal of the problem? Input ` 3 3 ` , Should have the output 0 1 2 , but they gave just 2

@AjayMishra 0 is not included

It is for everycase

0 is the output of first test case.

How one can conclude that it should not be included?

OUTPUT is the number of numbers that can be formed.

@AjayMishra It means every number can be formed

No such number(0) exist which cannot be formed

and what about the next one?

The number which cannot be formed are (1,2). Hence Output is 2

Which is the number of numbers(1,2) which cannot be formed

Output is number of numbers than can be formed and that cannot be formed. This is inconsistent.

Yea

@AjayMishra Sorry that cannot be formed

Oh.

Confirmed !! : that cannot be formed

@AjayMishra Is this chat.stackexchange.com/transcript/message/50675474#50675474

correct?

Ya.

What if n= 3 and k =3

That works.

@cattt only problem with that is, it counts one more.

That can fit into really (4-5 lines program)

Does that also include numbers less that k

Try this: Solve by hand when $k = 6, n = 3$

@cattt code.sololearn.com/c6xjTOiKxrUd/#py

You can try it here

fine @cattt ?

@cattt It do check, for k = 6, n = 3 , the output would be 11 , which are 1 2 3 4 5 9 10 11 17 18

although it requires a condition check when q = 0 , so that it print just k-1

I correct. Alpha = cieling (k-1/n-1)

And if you have to only count the numbers and not print them, it's even easier.

There was a small issue with last alpha.

@Mann it works even without correction, just one to tweak it a little bit.

Oh okayy. The problem was that there was some case where the code didn't run before. With the new alpha it does I think

Then it's good anyway

If it's just about counting the numbers then you can even get a formula for that

@Mann the program still won’t run good when k< n-1 , but one can make it work by case checking

No no, with new alpha it will I think

Just check.

I = 0 to alpha

I = 1*

I = 1 to alpha and the same number check algorithm

I think it should work

I just woke up, need to get fresh quickly and go to lab

pastebin.com/p62kMVRB

Here is my C++ code to find the number of numbers

Whats wrong in it

@Mann u see?

U there?*

I have used $k+n-1-i*n$

@Mann @AjayMishra

@Dante I will get back to you in a bit. Surely

I have seen your question

@cattt use this instead.

Oh OK. Let me know when you're free. Gotta ask some stuff related to admission also.

i*k - (i-1) *(n+k-1) -1

@Mann Which

And sum over all i from 1 to alpha.

Should work I think but check any errors see you

@Mann Including alpha ?

@Mann Its still wrong

@cattt I really don't have idea. I am almost sure it should work

@Dante hey

@Mann I wasn't in my place at that time. But alpha = ceil((k-1)/(n-1)) and alpha = floor(k/(n-1)) produces the same output. For you help: Floor() = GIF, ceil() = LIF

They product the same? It's slightly different for some values like k=6 n=5 I think

I'm uploading the screenshots, wait.

And ceil one is correct as far as I think

for k=6 and n=5

Check manually

Try k=6 n=5

I've included in, how range works.

okay. I will manually and then via program.

No just the program

first

It spits out wrong output only for first continuous patch of natural numbers.

Can you remove that +1 in range, I know it shouldn't be the case but Idk sure why

In the second for loop

Write ik and not ik+1

Just a test

yeah, works.

worked?

` 1 2 3 4 5 11 `

Yeaaaa

This should have been the answer exactly

Now I don't know why is Catt not able to validate

oh, that was my mistake!

I think this should work for the problem

Last number was not to be included

Got it got it

Infact if it's about counting these numbers only. You don't even need a program you know

It's just you can directly get an analytical expression

It works for the problem which I previously thought needs cases.

yeaa this seems perfect to me atm.

Btw for the algo, should I tell the approach?

It's a good exercise to prove atleast "intuitively"

No, I will put on the shelf where I use to put most things that need a good thought.

Okayyy

This is definitely a good problem

Try it later

i gotta go to market and buy stuffs and grocery and go back to my room. see you later

@cattt It is not clear to me what your algorithm is. There are too many variables and I never worked with that library which you've included. You can have 8 line of simple code, rather than 50 lines of code that uses modular arithmetic. I guess, that hurts your originality(assuming you come with your code yourself) but sometimes a tradeoff can be rationalized.

But why are using such huge numbers? program time limit is 1 sec[ Although useless don't consider whether I am on simple phone or using cloud to compute stuff]

@Mann Hi

Hello, koi mujhe please yeh batao ki counselling ke wakt first round ke baad reporting karna hota hai usme hume date milti hai kya kis din report karna hai ya 5 din mei kabhi bhi ja skte hai?

Mann had posted his whatsapp number here

Had anyone saved it?

@PolarBear Closing date to add choices for mock counselling ?

@Dante See website.

21 June for 1st one.

k

Haii

I am back

@Dante

Hey

Mind sharing your number again?

9888450502

but it won't work right now for calling purposes

Only whatsapp

Np, that's what I wanted

Oh okayyy then it's there