comingsoon.cpp's blog

This problem i coined up by myself can pass as a Div4A right?
By comingsoon.cpp, history, 3 months ago, In English

I was trying to solve an exercise while studying basic probabilities and while solving the exercise, I thought of this problem which I thought was fun, so I decided to share it with my fellow noobs :)

Please if you're anything above a newbie, (in Ercole Visconti's voice) This problem is not for you (no offence) :)

Problem Statement

You have n tiles numbered from 1 through to n. You are also given an integer K. The task it to count all pairs of integers (a,b) from the n tiles whose difference is K. that means (a — b) = K

Input The one and only line of the input contains two positive integer n and K. Output Print the number of such pairs (a, b) whose difference is K, thank you.

Sample input 100 11 Sample output 89

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3 months ago, # |
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n-k ?)

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    3 months ago, # ^ |
    ← Rev. 2   Vote: I like it +8 Vote: I do not like it

    Lolzzzz, I'm such a dumb a**.... My actual solution was (n — (k+1)) + 1

    Didn't realize it was that simple honestly.

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3 months ago, # |
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max(n — k, 0) ?

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3 months ago, # |
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I think that is good for problem A of Div4.

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3 months ago, # |
  Vote: I like it +75 Vote: I do not like it

Answer = $$$\max(\frac{(n-k)^2}{n-k}, \lim_{x\rightarrow\infty}\frac{3x}{2x^2})$$$

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