ENG  RUSTimus Online Judge
Online Judge
Problems
Authors
Online contests
About Online Judge
Frequently asked questions
Site news
Webboard
Links
Problem set
Submit solution
Judge status
Guide
Register
Update your info
Authors ranklist
Current contest
Scheduled contests
Past contests
Rules
back to board

Discussion of Problem 1264. Workdays

It is a very easy problem!!!!!!!!
Posted by XiangGuangTe 29 Apr 2004 11:05
The answer is "N*(M+1)"
Re: It is a very easy problem!!!!!!!!
Posted by CAHEK 22 Nov 2009 16:12
yeah, it's mega simple problem...
Re: It is a very easy problem!!!!!!!!
Posted by Beginner 30 Oct 2010 22:35
Yeah, it's simple but why my answer was wrong?
anyone can help?
Re: It is a very easy problem!!!!!!!!
Posted by Ivan Metelev 31 Jan 2015 13:34
Give me your code and I will help you

Edited by author 31.01.2015 13:34

Edited by author 31.01.2015 13:34
Re: It is a very easy problem!!!!!!!!
Posted by Rifat 6 Apr 2015 14:10
why "N*(M+1)" i can't understand pls help
Re: It is a very easy problem!!!!!!!!
Posted by Ealham Al Musabbir 1 May 2015 10:09
"For each integer from 0 to M the function would calculate how many times this number appears in the N-element array..."

Let's say N=2, so you have two-element array.
Let's M=4, so you have to check whether there is any 0,1,2,3 or 4 (since from 0 to M) in each array...
So total lines = 2*(4+1)
Hope, it helped.
Re: It is a very easy problem!!!!!!!!
Posted by Fardayek 29 Aug 2016 16:13
how could 3 or 4 in the n? cause n = 2 so 3 or 4 are not in the array....for this reason why we don't use if else statement to see whether m is bigger than n or not...but we calculate it straightly?...can you please describe?
Re: It is a very easy problem!!!!!!!!
Posted by liquid 11 Sep 2016 15:12
because, a box is given as a clue .....
and you have to solve the problem according to that.. i guess :/
Re: It is a very easy problem!!!!!!!!
Posted by Shipan Mazumder 5 Nov 2019 09:06
Thanks, It helps me