programmingevan Does this challenge seem oddly easy to anyone else? Like a warmup exercise?
Unchained "Warmup" is an over estimation.
Omnikron13 You are not alone. As far as programming is concerned this is just I/O and a single addition, so... 'challenge' isn't the word I'd use to describe this. =P
chestaisokay This was very very easy . I hope there are some good challenges ahead .
c650Alpha I agree. I saw this problem, thought I was overthinking it, and then checked the Discussion to see if I was right.
_faker_ Exactly the same happpened with me!
aloksingh3007 same here !
RubberSideDown The solution to this challenge is an excellent, basic example of an application of the pigeonhole principle, which definitely Falls under the fundamentals category.
jpjaspreetmahal I am here just to read comments XD
AllisonP This should really be called Maximum Draws (because it's worst-case scenario). Minimum draw would always be 2.
lvsz_ I like the idea of a best worst case scenario.
aigookenzaki ahhmm ahh can you help me to this account cuase i dont understand the terms can you help me plzzz
whyzee Thank you. Thats what I thought too.
Debu_Basu Yeah exactly! It occured to me once, but then it was clear when I saw the sample I/O :)
aigookenzaki ahh i am new in this account i need help cuase i dont understand can you help me pls
Neon Simple 3 pairs : R,B,G Draw 1 : R1 Draw 2 : B1 Draw 3 : G1 Draw 4 {R2,B2,G2} any of the pick makes a pair with previously picked color :)
revernus Thanks, I was able to understand the problem only after reading your explanation.
gokuvsvegita lol
ultimusdrago lol
sakib_muhit Perfect... :D
chvnikhil good
codermehraj thanks
sankireddyvinee1 nice..
panditsaurav6 Just a request is there any way we can write it as permutation and combination problem.
foxtrot I'm surprised nobody mentioned the Pigeonhole Principle :-P
thakursaurabh98 That was the main concept behind the question but still simpler ways! :D
sachingadagi Am I the only one who did (pairs+ 1) ? :P
shunn789 No.
trendsetter37 No.
M4GNV5 No. ++ ?
badillosoft Yes, you're right :D
num of pairs plus one, because in the worst case you'll get off each one of the different pairs, thus, you should get one more.
Griffin_Teller i did that too.
wjm91 This one is really easy. If you have n-pairs of socks, that means that worst case you pick one sock from each pair. From there any sock you choose has to be a match.
So by the pidgeon hole principle we get the answer to just be n+1
Surprisingly basic to be honest
codespinach One line of Python 3:
[print(int(input()) + 1) for i in range(int(input()))]canhascodez Ruby
puts (1..gets.to_i).map {|_| gets.to_i.succ }
Alternately:
puts Array.new(gets.to_i) {|_| gets.to_i.succ }
Or:
gets.to_i.times { puts gets.to_i.succ }
The last is probably most idiomatic.
ajinkyapathak import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;
public class Solution {
public static void main(String[] args) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */ Scanner sc=new Scanner(System.in); int t=sc.nextInt(); while(t-->0) { int p=sc.nextInt(); int a=p+1; System.out.println(""+a); } }}
This Code Runs All Test Cases...
arno_lee It takes more to reason about possible pitfals, because you're like "it cannot be that easy!", the level should be marked as Moderate
agusakov Easy. Pigeonhole Principle.
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