UPD: (28 April, 2022)
আবার চলে এসেছি! (That's Bengali for "I am back! (in Terminator mode)")
I am super excited to invite you to participate in Codeforces Round 752 (Div. 1) and Codeforces Round 752 (Div. 2) which will be held on Oct/30/2021 17:35 (Moscow time). This round is rated for both divisions.
You will be given $$$6$$$ problems in each division and $$$2$$$ hours to solve them. All the problems are authored and prepared by me.
I would like to thank -
- antontrygubO_o for his breathtaking(
literally) coordination of the round. no funny text this time - Alpha_Q, Anachor and antontrygubO_o for putting up with my dumb ranting and helping with the problems.
- Um_nik, gamegame, 244mhq, dorijanlendvaj, upobir, _Ash__, Anachor, Arg_007, Alpha_Q, steinum, Tasdid, t17, imAnik, Alfeh, Aritra741, SajidZakaria, YoyOyoYOy000y000, SlowDecay, ijxjdjd, flamestorm, ovis96, 16204, and Urvuk3 for testing the round.
- dorijanlendvaj for providing a test case which will potentially prevent few wrong solutions from getting AC.
- The great MikeMirzayanov for the amazing Codeforces and Polygon platforms.
- You, for existing and participating in the round. Also, thanks for giving your best and not quitting. You will eventually be happy
. Just hold right there.
The statements are short and directly ask you what to do and I have tried to make the pretests strong. I highly encourage you to read all the problems.
Solve problems and help the poor! Yes, you heard it right, this time you can help the world by just solving problems. I will donate money based on the solve count of each problem by the following measure:
The total estimated money is half of what I will get from Codeforces for this contest(but you can make it more by just solving more problems!). I am a student, so pardon me if that's too little.
Also, did you know that you can still upvote my The Ultimate Topic List blog and make it to the top?
Finally, I would like to dedicate this contest to me! I want to thank me for believing in me, for doing all this hard work, for trying to do more right than wrong. I want to thank me for just being me at all times.
Score distribution:
Div.1: $$$750-1000-1750-2500- 3500-3750$$$.
Div.2: $$$500-1000-1750-2000-2750-3500$$$.
Good Luck! Love you .
UPD: Editorial
UPD: Congratulations to the winners.
Div.1:
Div.2:
UPD: Here is how the donation amount looks like:
FYI I will get 400 USD for this contest. So the donation amount is indeed half of this as I have estimated before the contest, although didn't think that it would be this accurate.
I will update you again when I get the money from Codeforces, more likely after a few months as I haven't received my previous contest's payment yet which happened 3 months ago!
Also, thanks for being a part of this. You have just helped someone who is in need (I mean after I distribute the money).
Donation per AC!! Just loved it. Proud of you vai ♥
Exactly bro ! I just got surprised that how this man comes with such a amazing idea.
He should reveal how and where he's donating the money. It's just as important to know that the money will be utilized in the best possible manner.
https://www.facebook.com/100007930489461/posts/3203515616589423/
no funny text this time? :P
"0.005 USD" WOW THANKS MR.GENEROUS
I see your negative comments almost in every recent blog posts. Seems like you have opened this account just to pass rude comments!
If someone can help the poor just by solving problem, that's gonna be more fun!! :D
Yet another great round is coming ! :D
Great Initiative
That's great Shohag. Hope everyone will enjoy this contest.
Wow, that's great. Hats off to you.
Proud of you, bhai. <3
Super excited for a Bangladeshi round again!! :D
Very Good idea for helping poor people. Thanks for this contest. I am so happy that I can help poor people.
Hope for 2K upvotes and Rank #1 in top contributors.
Now his contribution becomes the same as 1-gon lol. Looking forward to seeing him become the top.
Update:Now he is the top.
This is what I get for stating facts. :(
:(
Inviting the fellow programmers to be the part of this initiative....
Let's send YouKn0wWho vai(brother) to the Mars by upvoting.
SecondThread once told to send 1-gon to the moon by upvoting. Trying to copy him. :) https://www.youtube.com/watch?v=FsQPMmRhUGA
এভাবেই বারবার চলে আসবেন ভাই ♥♥
Very excited to praticipate of a bangladeshi round.
Donation per AC!! Too good to be true.
Again , pari na pari khela hobe :3 (it means i will must participate ,dont downvote)
Your last round was extremely good. I hope this one is as well.
That was a tough Math round
how many did you solve or you like the testcases
For a moment I thought why the blog has only 166 upvotes(previous round announcement has more than 1600) then I realized that you are back with a new round .
Really looking forward for this one after the last round. Also props on the idea for donations <3
Super excited Vaia. Take Love.
Codeforces Round #752, writer: YouKn0wWho
The meme has come true
As a tester, these are some of the coolest problems I've encountered in quite some time. So if you're going to participate, you're in for a treat!
as a tester, I would say the problems seemed very cool to me. don't dare to miss the contest and the potential high rating! Good LUCK all!
just curious.. how much does codeforces pay one for setting up a contest?
look at this
This contest is Div1+2 then CF will pay 400 USD.
Actually i was trying to estimate the total amount and it came up around 350+ dollars. Damn, i should have seen this comment from the start.
Orz YouKn0wWho
Hyped for more anime references in the questions lol
So if I participate from Div1 and solve all problems, you donate 6.255 USD, that's nice. But that's too tough, I'll just donate 7 USD myself instead.
I thought you stopped being a nerd.
Or better.. I'll donate 7 USD, so give me solutions to submit during contests. :catThink:
Okay this is bad, now we're plane selling and buying solutions.
(give me some I'll give you 14 USD)
Deleted
good to see u back
I don't have any say over whether it will be in pretests, YouKn0wWho just asked "Is there a test case that accomplishes [REDACTED]?" and I responded with a test case that does.
Right! i wasn't sure who had a say. I've updated the image now :P
Just out of curiosity, What test case of which problem it was?
It was a test case for long long overflow if modulo is only taken at the very end in div1C.
Eagerly Waiting
CHAD YouKn0wWho
Hello everyone. Good luck to everyone on the contest and a good rating!
Finally, I would like to dedicate my participation in this contest to myself, for doing all this hard work, for trying to get to Expert. I'm just so grateful to myself for being me at all times.
<3
♥ ♥ ♥
osthir vai <3
Positivity Everywhere!!!
This is the first time cheaters will actually do something good to this community lol
but what happens if someone leaks div2F and 1000 people copy it
the good deeds
Their excuse would be "I just want to help and give back to the community, so we did this for a good cause so please forgive us Mr. Mike"
Masha Allah
Last contest by YouKn0wWho, I gained 125 points. To date, it was my best contest ever.
Update: Candidate Master this time
Congrats on CM!
Should we be concerned that the author has a monetary incentive to make weak pretests?
Nah I think we should be more concerned that your rule might be coming to an end.
I guess if he had any monetary incentives, he would prefer not to introduce a donation system at all
That's not fair. The donation system is the political move that dethroned 1-gon!
Should we be concerned my words have come true?
Yes, we should be celebrating the new ruler.
I honestly can't tell if you are being humble or trying to get upvotes. Let's stick with the former.
Thanks for the generosity and look forward to this round!
Hope I can donate more than 0.02 in this round :D
I hope this will be super exciting contest like here
Deleted
False, IIT BHU had previous div1 rounds
Ashishgup is not the authors of the round you mentioned
yup sadly he dont but in the upcoming days he will have :) , anyways still false manthan by IIT Bhu was div1 rated in past.
Okay, after reading the full passage, i am getting more positive vibes about the contest.
Looking forward for a great contest and a motivation to keep me going.
Poor-force will help me come to master this time
Monogon are you scared of YouKn0wWho??
অপেক্ষায় আছি, ভাই।
Second time participated in your round! Love it.
YouKn0wWho is one of the best setters I know in Bangladesh.
Again 1k+ upvote Sohag brother ........ <3 Keep ahead
Hopefully #752 will be interesting
Yes
Good luck too all :)
You mean upvote more? coz u shdn't participate with multiple accounts.
Good point
no, I meant Solve one problem multiple times and help the poor. just kidding... I know the rules...
That's quite an unique idea! At least for me this is my first time to participate in such contest.
Hope for a great contest!
Why downvotes to his comment?
I also don't see anything wrong in the comment.
Maybe because EasonCF's comments are everywhere and people got fed up with them
Exactly
But I don't think this can be the reason to downvote a comment.
I think whether to downvote a comment depends to itself but not depends to the author.
It seems like saying stereotyped words to get contribution but not expressing his/her own words.
Oh,that should be downvoted
Appropriate.
Where will you be donating the money to?
Solve problems and help the poor! <3
As a tester, I recommend the contest to you all. As a contestant upvote the blog, or else ...
The statements are short and directly ask you what to do. Thanks for assure it and not to make the statement gazakhori and azaira pechal. :D
Thank you YouKn0wWho for keeping the problem direct and short. And your previous blog is helping a lot. ^_^
Apart from the contest it will be more exciting to see YouKn0wWho as the number one contributor ;D
number theory — "I am back too" xD btw great initiative ❤️
"Donation Per AC". Go ahead with great work vai.
Upd:
Div2B: 0.005 -> 0.006
Div2C: 0.005 -> 0.008
Div1F: 4 -> 2
I hope now the distribution is more consistent with the point values of the problems.
I want to become Grand Master!
me too bro
Great initiative, you are doing a very good job. Thank you!
According to the rating predictor for last contest (education 116), my rating will go up to 1905. That means this will be my first time to participate div 1. So excited.
it's really good stuff that you are doing with this job, respect!
My solution for B here, https://mirror.codeforces.com/contest/1606/submission/133495767, fails in test 6. It appears to be some rounding errors. Does anyone know what caused it?
I want to take part in this contest but tonight, Liverpool — my favourite football cbub have a match wwith Brighton :(((
I gave contest on the day pakistan beat india for first time in cricket world cups in 29 years . so deal with it .
let me think, I think I can watch the first 30 mins and after that, I join in the contest =)))
অনেকদিন পর আজকে পার্টিসিপেট করবো ভাই :) ভাবতেই ভাল্লাগছে...
Looking forward to my first Div 1 round!!
Alhamdulillah.. Great achievement! YouKn0wWho is now top contributor from Bangladesh. Proud of you man!
It is good round!!!
Excited for this one as it`s gonna be my first contest on this platform!! Looking forward to gaining some nyc ratings .....
As a Bangladeshi Participants, proud of you bro
Congrats on becoming the new top contributor! Remember, great power comes with great responsibility.
If I'll solve E/F and resubmit it 9 more times, will you donate 20 USD?
But they have to pass the main tests and sadly CF skips multiple AC submissions for each problem :((
This alone made the contest worthwhile YouKn0wWho
I locked one of my codes for hacking but I cannot view the locked solutions of other users in my room. Kindly fix it YouKn0wWho
I also faced same problem :(
I wish, I also had the same problem. In the last 1 minute, I tried to hack someone's solution and my rank went from 86 to 126.
Very interesting problems! Thanks to creaters of this contest!
but very strange problem C and I dont know how my solution pass pretests
Great problemset, but div2-D was quite easy compared to 2000 rated problems i think.
D was too easy tbh. C made me realize how dumb I am once again. Nice contest tho
Was it? I spent over an hour staring at it and couldn't come up with solution xD
I think he is talking about div.2 C and D:)
I am pretty sure natofp is talking about Div1B (Div2D) based on his submission time lol.
Took me an hour to do it :|
What approach did you used for D?
Did bruteforce and found the pattern.
To add to this, a little bit more explanation on how I personally came up with $$$y-(y\mod x)/2$$$:
Suppose $$$x\leq y$$$. We want to find a $$$n$$$ such that there exists integers $$$k,l,b$$$ that satisfy $$$n=kx+b$$$ and $$$y=ln+b$$$ (and $$$b \lt x,b \lt n$$$).
We can see that any valid $$$n$$$ has to satisfy $$$x\leq n\leq y$$$ (*), so $$$k,l\geq 1$$$.
Expanding $$$y=ln+b$$$, we get $$$y=(lk)x+(l+1)b$$$. This means $$$(l+1)b=y\mod x+ax$$$ for some integer $$$a$$$.
To construct a valid solution from here, note that since $$$x,y$$$ are both even, we can set $$$a=0$$$ and $$$l=1$$$ (since RHS is also even). This gives in a valid $$$b$$$ and $$$n$$$, which is exactly $$$n=y-(y\mod x)/2$$$.
(*) To elaborate on why $$$x\leq n\leq y$$$: If $$$n \lt x$$$, $$$n\mod x=n$$$ but $$$y\mod n$$$ can't be $$$n$$$. If $$$n \gt y$$$, $$$y\mod n=y\geq x$$$ so it can't be $$$n\mod x$$$.
I found the pattern after bruteforcing testcases, dont know why it works.
basically if x>=y, ans is x+y, else y-(y%x)/2
Maybe I just happened to figure out the pattern
Thanks for great contest! I was late for an hour because of my classes but still participated in order to help charity!
WOWW.. You might not know it, But man you are a hero.
Countforces Round #752
Problem 1604C - Di-visible Confusion is logical problem ...... I didn't solve this.
Did bruteforce and found the pattern.
How to get rid of TLE in pretest 5 in div1C?
I needed to get the possible next numbers can appear after index i and their count also. To do that, I needed to maintain a set.
133677385
Why a lot of solves for D div2 -__- ?
D is a one liner solution which is perfect for cheating
Why do I think $$$O(n\sqrt n\log n)$$$ can pass 1e5 in 4s？
It can pass. But set runs too slow that I have to replace it with array+sort :(
Actually $$$ \log n $$$ is unnecessary
We Only need $$$ O( n \sqrt{n}) $$$ (Assume $$$n = \max a_i$$$)
Lol now I feel dumb... I used a seg tree to query on $$$O(\sqrt{n})$$$ different intervals, of course I didn't realize they will all go to the same state in the next step so there are at most $$$O(\sqrt{n})$$$ different states at any step lmao.
Good Contest , I misread D to be find number of n from 1 to 2e18 which satisfy that condition , in the last 20 mins after looking at number of submissions I read it again , felt horrible and then mind stopped working . Anyways good contest man .
How to solve C?
Let me know if you find out about approach for C. Managed to solve div2 D but couldn't solve C :(
you can iterate over array and count number of items that mod(i+1)!=0
moreover at each step you should include items that mod(i+1-1..current_count)!=0
if at some step you found items that mod(i+1-1..current_count)==0 then answer is NO otherwise YES
the problem is saying check if at least one number from 2 to i+1 isnt a divisor of a[i]. Its enough to check from 2 to min(i+1, 40) cuz lcm of 1...40(or so) is too big for a[i] to fit
This is so because for the current number, if any number before it cannot be removed, ans is No. Else, you can remove some numbers and check if current number is divisible by i+1-count of previous numbers removed which results in the range 2 to i+1(notice removing numbers after current one does not affect the position so its pointless for current number)
The problems were very interesting. I felt dumb after getting the idea of every solution I could xD
Can someone tell me what I did wrong in C: If there exists no element a[i] which is a product of all the numbers (i+1)*(i)*(i-1)...2, then we should output YES. Is this correct?
I think it should be least common multiple, i.e, it shouldn't be 2 * 3 * 4 == 24. It should be the LCM of 2, 3, 4 which is 12.
you should check lcm not product.
$$$q$$$-binomial? Are you serious?
Sadly we didn't know about this. But it's true that we were afraid that some
chinseguy would use someweirdstuff to solve the problem. I hope you liked the other problems.I got the idea of Div2D 10 minutes after the contest. I hate my life
How to solve div2 C? Managed to solve div2 D but blundered in C :|
Nice contest. But lots of maths (:
Didn't really like Div2 D. It's a very simple solution, and either you find it very quickly or you sit around thinking for a long time before you realize it and bang your head
Yea, too me a whole lot of time. At the end it was a simple pattern lol. Did bruteforce and found the pattern.
How to solve D ?
Did bruteforce and found the pattern.
i think:
if(x==y) ans=x if(x>y) ans=x+y if(x<y) ans=y-(y%x)/2
although i couldn't submit in time..
if y < x then u can choose any n larger the y so that n % x = y holds
otherwise if y >= x and suppose y % x = z, then if you find the highest multiple of x which is smaller then y and add z/2 to it, n will become evenly match for both numbers meaning n % x will equal to y % n.
I think this time zhoukangyang will become LGM finally!
And is he the youngest LGM in codeforces?
How old is he? less than 16?
less than 14
Can someone give some hints to div1 E . I only have a naive $$$O(n^6)$$$ solution .
I have the O(n**6) solution but only traversing states which are reachable from the beginning and from which the end is reachable, and apparently it's fast enough. Although I guess it depends on which exact states your solution has.
My solution:
First assume that the sequence is nondecreasing . Then enumerate the first number and do a dp , which record the prefix sum and the last number .
In my case the difference is that after fixing the first number I put numbers from big to small, and therefore my state is what is the maximum allowed number, how many numbers we still need to put, and what is their maximum allowed sum.
So I guess having "maximum allowed sum" when going from big to small has fewer reachable states that having "sum" when going from small to big.
My states are choosing $$$i$$$ numbers between $$$[l,r]$$$ and the smallest number is $$$l$$$, the sum of the numbers is $$$s$$$.
How could I improve?
Why is the time limit in D so tight??
I have a solution that runs in around 3.4s locally, and is $$$\mathcal{O}(n \log^3 n)$$$ except for a $$$\mathcal{O}(n \sqrt n)$$$ part that takes just 0.5s. I'm sure I could make this pass if I optimised my segment tree more, but the problem shouldn't require that!
EDIT: optimised the segment tree to this special case, and now it passes (133687383). Here's a submission with my usual segment tree code (133688464).
Is it possible to do problem C in n^2.
TLE
Dear YouKn0wWho Your contets are very difficult for me T_T
That's what contests are for, practising and competing and having some fun. Solve some difficult problems rather than solving 200 problems of 800 difficulty.
What approaches were supposed to fit into TL and ML in Div1C by the author, and which were not supposed?
lol. The final pattern was a simple 3 line code.
Lol, I think my Div 2D code amassed to a couple of lines (really only 3 "important" lines of code). Out of curiosity, did the code work?
long long n=55; int m=20; if(m>n){ //code } Is it valid to use comparison operator between long long and int data type?
it will first convert int to long long before comparison, so, yes
Here is some feedback on the problems:
Overall, the contest was very well-prepared and it gave me some strong emotions, so thanks to the authors. The difficulty distribution was perfect. I am eager to read the editorial to find out how to solve D properly.
Can you elaborate on the time complexity of your D? Editorial also mentions D&C and states that its actually $$$O(n \log^2 n)$$$, which I'm a bit concerned about (more like $$$O(n \log n \sqrt n)$$$).
The cost of an interval $$$[l, r]$$$ is given by ($$$q(x)$$$ is a function we have precomputed, namely the number of coprime pairs $$$1 \le a \le b\le x$$$)
I think that your doubt on the complexity comes from the fact that computing $$$c(l, r)$$$ may require $$$O(\sqrt{n})$$$ operations (if this is not the issue, please clarify why you think that the complexity might be $$$O(n\log n\sqrt n)$$$.
The trick is that, with some precomputation, we can compute $$$c(l, r)$$$ in $$$O(1)$$$. We precompute, for every $$$r$$$, the values $$$c(r/k, r)$$$; this can be done in $$$O(n\sqrt n)$$$. Then, we have
where $$$m$$$ is the smallest number such that $$$\lfloor \frac{r}{m}\rfloor \gt \lfloor \frac rl \rfloor$$$ (it can be found in $$$O(1)$$$).
Since the cost function can be computed in $$$O(1)$$$, and we are executing the D&C dp optimization $$$\log(n)$$$ times, the total complexity is $$$O(n\log^2(n))$$$ (plus $$$O(n\sqrt n)$$$ precomputation).
Sorry for confusion, I didn't read the editorial, which apparently has the same solution as yours.
You understood me correctly, the end of editorial mentions solution that uses $$$O(\sqrt n)$$$ per one query. It's now edited, stating that in practice it fits in TL but no proofs provided on the complexity (I think it's still $$$O(n \sqrt n \log n)$$$).
Btw, I've used this technique like 2 times in recent days, but wasn't able to apply it here, what a shame...
in the problem Div2D, why x < y : cout<< y-(y%x)/2<<"\n" ?
If you bruteforce for the values of x and y, it was following a pattern.
Can someone prove it with math?
For x < y, lets take K for which, K*x < y < (K+1)*x.
The N for this case will be , (K*x + y)/2.
N = K*x + (y — K*x)/2
N % x => (y — K*x)/2
y % N will be y — N (Since 2*N > y)
y % N => (y — K*x)/2
if x < y, then suppose y % x = z, then if we find the highest multiple of x which is smaller then y and add z/2 to it.. then n % x = z/2 and y % n = z/2, cz they both are even numbers and I can find a midpoint where both numbers multiples distance is same.
Nothing is better than doing charity while having an interesting competition!
Unpopular opinion: Div2 was only observation and math, no programming. Boring experience.
Sure, but isn't 'observation' key to 'problem solving' which is pretty much what competitive programming is. A keen observer is usually also a shrewd problem solver.
No, but I expected more data structures, algorithm concepts such as binary search.
Implicitly, in your argument, you assume that observation means math (or that, observational problems are inherently mathematical problems). I don't think that's true. Sometimes just observations can come from realizing that one needs a certain data structure, like recognizing the application of segment within a particular context.
I agree that observation is key to problem-solving, but I don't think this justifies having such a mathematically heavy contest.
Does the writer have fetish with XOR? Why so many XOR problems from the same writers?
Can someone explain why this code gets WA? I literally did the same thing as explained in the editorial Link: Your text to link here...
You don't read the complete input before printing the output for a test case. Then when you'd try to read the next test, you'd first receive the input left from the previous one.
Yeah I realized that in the meantime. Stupid of me, what can I say. Thanks for explaining anyway!
You don't need to check for all the n elements, just for the first 21 elements.
.
I just went ahead and downvoted you for saying that. Thank you
Just solved Div2 D just 8 sec before the Contest Ends, Love my Life and codeforces
It was a math contest, didn't even need computer.
There were so many $$$69$$$'s. YouKn0wWho surely loves $$$69$$$.
I want to know the name of the man who ranked 2 during the contest and suddently dissappear after system test .-.
anyone calculated the final donation amount?
is this contest unrated?
it is rated ..y did u suspect that this is unrated?
My personal feeling — E is much, much easier than D. For me E is a pretty standard task like "you have an obvious DP with huge complexity, optimize it with some observations and harmonic series principle". D's idea is obvious to me too, but implementation and some specific details make it harder for me.
Despite my feelings of losing rating and my contribution throne, the problems div1ABC are pretty nice. If I have a suggestion for improvement, it's to have more topic balance. I thought there is too much number theory. (As much as integer division and remainder can be called number theory)
Great contest I enjoyed solving the questions and also happy for donating.
Silly thing happened to me today, I forgot to save file after removing cerr line and submitted the unsaved file. It passed pretests but TLEd on actual tests. Is there any way to use fast output with cerr?
P.S. use clog instead of cerr
The contest was good, though i think it was a bit mathematically. thanks the writers
Does anyone know why Itst_boyfriend was removed from the contest?
When will ratings update?
Can any one make me clear how the O(N²) solution is getting ac in div 2 question C???
Brute force works here because it is not really O(n^2), because actually most numbers can be removed in most positions. So a linear search for such a position does not take n steps but at most aprox 20, and on average about two or three. submission
This guy never disappoints. ❤
I mistake
you know it's a bad contest when problem D has O(1) solution
Dear programmers, can you help me? Recently, for fun, I created a team of friends at Codeforces. After the contest, Codeforces sent me an email saying that my solutions are very similar to the solutions of alex_2008, which is in our team. The fact is that I am also alex_2008. And there in the letter it was said that for repeated violations of the rules, they would block my account. Can you tell me what to do?
I participated in the round yesterday. I received a mail after the contest that my solution for DiV2 Problem B coincided significantly with two more solutions(My solution -world_classic/133631831 133631831 others solution-keetzo/133671951 133671951 , kaypee284/133672077 133672077) . As the question had a straightforward solution and the other users coincidently may have the same templates as mine so it may have coincided by chance. I request MikeMirzayanov and the team to look into the matter...
I am amazed at how accurate your estimation is!!
As a tester, hope you got nice rating
wrong place
So I promised that I would donate a certain amount of money based on the number of ACs that each problem gets. The idea was to motivate people to solve problems and at the same time help out real people by just having green verdicts.
The amount turned out to be around 200 USD (I have received 400 USD in total for that contest).
So following that event, today I have distributed Iftar platters (Ramadan Meal) of the same amount among 120 rickshaw pullers and poor people in general. Thanks to Toushiful Ferdous Badhan and Rakib Hasan for helping me out.
Posting here to let you guys know that, hey, your green verdicts helped someone to have a green moment. Also, you solved a div2A problem but that way you just helped someone solve a real div1F problem which is Food (may it be for one day).
We can’t change the world overnight but we can play our parts, may it be very little! I like the analogy of one of my friends which she calls "The Wave Analogy". The particles stay where they are, they just dance and in consequence, a big wave is formed!