Good day, Codeforces! Or in my language, magandang araw, Codeforces!
Codeforces Round 597 (Div. 2) will be held on Nov/01/2019 17:35 (Moscow time). Is it rated? Yes, but only for participants below 2100 rating.
Of course, this round is not made by me alone. So, I would like to thank the following people for their help in making this round possible:
- coordinator awoo, for coordinating my round even when he was very busy;
- testers antontrygubO_o, McDic, 300iq, adedalic, isaf27, farmersrice, 265918, vovuh, CaseRuten,Stresshoover, nooinenoojno, Pavlova, defolaut, and Arpa, for testing problems and giving very valuable feedback;
- and of course, MikeMirzayanov and Codeforces staff, for providing us with the amazing Codeforces and Polygon platforms!
There will be 6 problems, and you will be given 2 hours to solve them.
The scoring distribution is as follows: 750-750-1250-1750-2250-2500.
I hope you enjoy the round!
Update: editorial is out now
best round i ever tested. keima orz
Have you forgotten that you were testing my round
best round I ever tested
worst round I ever tested
worst lie you ever lied
they just forgot to insert me :)
best thing they don't insert you .
NO
keima orz
I'm pretty sure it will be another bad contest. If you agree, upvote me.
best round i didnt tested
r/technicallythetruth
Your comment gave me goosebumps.
*tasted
Are you the first one from Philippines to make a contest on codeforces ?
No. robinyu made Codeforces Round #419, https://codeforces.me/blog/entry/52653.
I want to AK it,but I can't.
I hope I can become candidate master after this round :DDDDDDDDD
I hope, that this round will be balanced, and all people get results, they needed.
After almost 3 years I am back on CodeForces and gonna participate in this round. Wish me luck!
You lied. You didn't participate in this round.
It clashes with a codechef round , but after reading comments from testers ,it seems like, i should ditch that and participate here ...
best choice ever made
Same here I also ditched codechefs round and got +83 here
I think it must be one of the best contests in Codeforces!
I wish contest has strong pretests. I don't like to take wrong answers on main tests.
Whenever 300iq was taster , 193 tastcase passed and then suddenly wrong answer in tastcase 194
As a tester, I enjoyed solving problems. Well written problem statements mixed with Chinese culture. One of the problems is about a childhood game and it was completely interesting for me!
how interesting would it be a D or E problem about game theory :P
It'll be interesting if C was combinatorics :P
I hope it is not interactive, I had bad luck trying to interact with my Chinese friends' kids!
Keima915 OP, from green to orange in 7 months .. interested about the contest to start my 7 months journey :p
I think it's not that hard. Good luck!
hope solve a,b,c
Please can anyone say me this round I wanna be specialist how many problems must I solve
Like what you did in your last round.
How can I improve? advise me something.
I already participated 27 contests result is newbie.
Always getting out of your comfort zone. And focusing on fixing your mistakes.
I think you only need solve 2 problems in short time or 3 problems
How they calculate it to be specialist?
By your relative ranking to other people, and your previous rating vs. other people's ratings around your ranking
huh.. very complicated, i will go and change the css to red.
Expect for the constest!
want for the contest!
I wish I can get expert in the contest again. :)
Good luck to everyone!
I m back Bois.. Had a busy October but now I am back to top the charts again.
The contestants had to face some difficulty in the previous contest due to several server failure. Though it was overcome in few seconds but its disappointing to reload the page over and over again during the contest hour. Div-2 rating changes also had to roll back because of some problems and we had to wait for several hours to get the rating changes in the last contest. Hope the authority will take care of these matters in this round.
Auto comment: topic has been updated by DeliciousFlatChest (previous revision, new revision, compare).
Best of luck in your first contest!
Why first and second question has same score distribution....does that mean they have same difficulty level??
Hope we have a fair race :)
I hope that I can turn it into a blue name.
Another DISGUSTING reading contest.
Reading is also a part of CP. So you should adjust yourself to read long questions. Else make the question look smaller to you. How? You figure it out ;).
I don't like to read tonns of text after hard work day. It tires the brain completely.
Do you think everyone is only doing CP? Are they not tired?. Let's not make any excuses and just read the question there is no other way.
Well, I guess that's just your excuse for not being able to solve problems on this contest. The statements were actually well written and short. Go practice instead of complaining.
Well, I don't know if the statements were well-written but problem E's statement is definitely not short.
You're right, they were well written except for E. When I saw that statement I didn't even try to read it and went for problem F.
strong pre test cases for problems at code forces I ever seen!!
Strongest pretest 13 I have ever seen
Wrong answer on pretest 13 on problem D, No idea why. :(
Any one have idea what is going on in pretest 13 ?
TC 9 of problem C?
Were u taking modulo when calculating the fibonacci ????
That test case got me too. I spent the last hour trying to fix whatever was wrong my fibonacci-number calculator.
I was modding by 10^9+7 everywhere I could see, but for some reason, my 89-th and onward Fib numbers were wrong. So of course the Fib products were also wrong too. Please send help
I think. You should replace ret*=fibNum(x); as ret = (ret * 1ll * fibNum(x)) % M and also temp = (temp1 * 1ll * temp2) % M;
Not sure what the || (or? or something else...) is meant to do or what difference ret*=a vs ret=ret*a would make, care to explain?
I think he means 1LL. LL makes it to long long
that is 1LL.
which leads to long long int.
Your solution will get AC after doing that.
Thank you so much for explaining that starboy_jb. I will never forget to cast again.
If you need help tell the approach of your solution rather than asking what was the test case. After some time you will get to know it right? :)
I found out that ways to arrange characters of length i which are consecutive, can be found out by this formula,
Pseudocode
In my solution, i was doing like this:
Is the problem with Modulo that i am doing? Please someone help.
First of all, your first "if" statement has identical branches.(now ok)
Secondly, dp[1] should be 1
Thirdly, by the formula you have given dp[4] = dp[3] * 2 + 1 = 7, but actually dp[4] = 5. For example :
intput:
uuuu
1.uuuu 2.wuu 3.uwu 4.uuw 5.ww
Here is my output of first 10 numbers. Check once again, i think you are misreading something.
dp[5] = 8 not 11
Then how I was able to pass 8 test cases? lol
till 8 TC,
the continuous occurrence of n and u is not greater than 4.
Thanks anyways, I was calculating repeating cases again and again.
where 8,10 and 11 are repeating. Wasted 1 hour thinking about something wrong with computations.
Thank you :)
I calculated according to your first two lines of your comment (the if else part), you said dp[i] = dp[i-1] * 2 + 1 if i is even, maybe you have mistaken even with odd there, anyway, your formula is incorrect I guess.
how to solve D?
Kruskal or Prim to find smallest tree. However the graph must be modified a little bit. Add a new node (let's say it's the power supply for the power stations) and the edge from this source to each of n nodes is equal to the cost to build a station there.
Dont we need to make sure that there is atleast one power station? Wouldnt your approach fail when the power stations cost are really high, but the points are close by?
However, your minimum spanning tree will also need to include the new node, which represents a sort of "super power supply" and ensures that at least one node is connected to the new node, and thus has a power station.
Hmm? a tree is a connected graph, thus the newly added node (power supply) must belong to tree too. If the power supply is connected to node i, then you put a station at node i. Since the supply is always connected to some city because of connectedness, there is always at least one power station.
No because we are finding a spanning tree which is connected if the original graph is connected too thus there'll be a path from the power station to every city.
What the hell was pretest 13 in D :(((
Did you consider the case when $$$(x_i, y_i) = (x_j, y_j)$$$ even if $$$i!= j$$$. I think this is the mistake, I am not sure though, because I too got wrong answer on test case $$$13$$$. I didn't consider this case, so maybe this is the mistake I made.
IS it matters? Because dijkstra will take care of this fact I guess. CAn you please give the example by which you debug your code?
I considered this case. I was not finding MST though, which I think was the expected soln
approach for problem C?
I was getting some kind of fibonacci sequence
say we have len length of string of either n or u, in how many ways we can make pair in the string
so if we make x pairs of 'n' ,we are left with len-2*x 'n' characters and we have x+1 blocks where we can fill them , so distribute len-2*x objects among x+1 persons such that each get zero or more with (len-x,x) ways
and iterate x from 0 to len/2
Any chain of U's can be broken up in fib(length) ways:
Proof: if there are n of them: UU...U we can condition on whether to turn the first U into a W, or not: If we do, then it becomes W(U...U) (with n-2 U's) If we don't, then it becomes U(U...U) (with n-1 U's)
Now each k-chain of U's or N's in the string has Fib(k) possible strings, so we multiply Fib(k) over all the k's
Nice
But how did you calculate the Fibonacci sequence?
I was doing mod 1e9+7 during the calculation and I'm getting wrong answer on pretest 6
How did you come up with this intuition?
For my case...I understood this by observation, I calculated the results for the first few lengths of chain and then realized the pattern :3
Consider u as 1 and w as 2 , then how many possible combinations exist such that sum is n.
Let n = 4
uuuu = 1 + 1 + 1 + 1
wuu = 2 + 1 + 1
uwu = 1 + 2 + 1
uuw = 1 + 1 + 2
ww = 2 + 2
There exists 5 ways which is nothing but fib[4].
I saw it in a book (Proofs That Really Count) in the first chapter. It explained that fibonacci numbers can be thought of as the number of ways to tile a 1xn grid using (unlimited) dominoes of size 1x2 and 1x1. There you condition on whether you place a rectangle or square domino in the leftmost place.
what is test case 13 in D
I thought this was an amazing round :D What lovely well framed questions
Does problem D using Prime Algorithm? I did it but got WA on test case 4. What is TC 4 :(
I think you mean the Prim Algorithm. As far as I know, yes.
Yes, Prim Algorithm. Sadly, i still didn't solve it. After solving A, B, C for around 50 minutes. I came to problem D, the statement is so long and demotivate me a lot. After relaxing 10 mins, I decided to read the statement because I had 1 hour remaining time and the problem didn't hard as much as I think. If I started to solve it sooner, I would have more chances to debug. It's experiment for me. Very nice contest =)))
Yeah, I understand. I was puzzled too when I saw the super long statement. Just calm down and don't be afraid of those things. I believe you can do much better in the future. Good luck!
Slightly off-topic, but I love comments like this for the example they set for the CF/CP community. :')
I solved D with Prim's algo.
Dijkstra?I did using Dijkstra....But I guess the algorithm is similar XD
It's almost the same algorithm.
-is-this-fft- Is this prim?
Edit: I need to re-order these two statements for AC. lol
It seems so
hmm.. 13 is a devil's number i must say.
test cases should be 1..12 and then 14..30
Kruskal is fair enough
Check out if you have used "long long int" instead of "int". I also got WA at test case 4, and this helped me :)
I used long long instead of int, but It didn't solve. Btw, what is the difference between
long longandlong long int? are they same?Yes, they are identical.
My situation is exactly the same as yours.Long long didn't save me.(•́ω•̀ ٥)
Hey bro~I find a lil issue in my code,I defined a variable as int but it should be long long.Thus the program got overflow.Maybe you should check the type of each variable carefully.
Thanks bro, I definitely solve it again. I have just woken up.
I lack of dummy city.
What is pretest 13 of D? Wasted almost half hour :(
Try this:
Ans: 110
Try this test
UwU
dpforces
So true.
How to solve F?
Dp with digits (in this case is binary digits).
You have a+b = a xor b + a&b. Then a + b = a xor b if and only if a&b = 0.
I pre-solved it with digit dp. Let $$$f(pos, smaller_1, larger_1, smaller_2, larger_2)$$$ be the number of way to fill the first $$$pos$$$ binary bits such that [ the first number's prefix is already smaller than $$$r$$$ ], [ the first number's prefix is already larger than $$$l$$$ ], [ the second...], [ the second...] and their prefixes have no set bits in the same position so far. The transition is not hard to come up with. The answer is the sum of all entries in $$$f(0)$$$.
I am trying to understand your solution and transition. Could you please explain the transitions and the continue statements? I think this is a much generic approach. Thanks
PMed you :)
Could you please forward the message to me too. I am also confused as to what would be the transitions. Thanks in advance.
Could you please forward the message to me too ))
I finally realized that it was problem E until I submitted my code(since mirror sites don't show the indices of problems
I fooled myself XDDDDDD
Nice round. F is a fairly common problem btw. But it is still nice :)
Auto comment: topic has been updated by DeliciousFlatChest (previous revision, new revision, compare).
Nice Contest and problems.
But A was quite similar to this problem.
orz
Find out why this gets WA8 (takes less than 5 seconds), i couldn't during the whole contest :DDDDDDDD
Test: mnnn (should be impossible)
Question #2: which lines cause the trouble?
for i in range(1, len(s)): if s[i] in ["w", "m"]:
This doesn't check the first character!:^)
Answer to #2. Because the range of index starts from 0, not 1.
Ques.D) Shichikuji and Power Grid: In this question which concept and data structure are used. I think so Kruskal for the shortest path and DP for choosing the powerhouse or connection.
I used priority queue to solve.
First, push the cost of setting plants at each city into priority queue along with city number.
Then, start where cost of setting plant is smallest.
Then push the cost making a connection from this city to every unelectrified city (along with city details) and push them into priority queue. Extract minimum cost (setting plant or making a connection) and go to step 3 (n-1 times).
okay cool, a different approach
it sounds like you just invented Prim algorithm lol
My solution: Using Kruskal:
Create all the possible edges and their weights(using formula in the statement). Sort the edges according to their weights.
Suppose you are trying to connect the connected component U and the connected component V using the edges (x, y). Basically you have 2 options:
Build a power station in a city in U and a power station in a city in V (Just pick the city with less cost). You don't have to build the edge (x, y) anymore.
If you decide to build the edge (x, y) to connect U and V, then you have to build a power station in a city in the union of U and V. Just pick the smallest one.
Just pick the optimal options for each edge (x, y) and add up the cost.
In A I guessed from test cases that a and b needs to be coprime, and the submission worked, But I don't get why is this correct answer. For example a = 3, b = 7 there are infinitely many numbers which are coprime to 3 as well as 7, then aren't there infinite black numbers?
Check this: https://en.m.wikipedia.org/wiki/Coin_problem
A number $$$X$$$ will have color white iff $$$X$$$ can be written as a combination of $$$(a,b)$$$
==> There exists $$$k_1,k_2 \gt =0$$$ such that $$$X=k_1*a+k_2*b$$$.
So the numbers $$$X$$$ having color black will be the ones such that that equation has no solutions. Since that's the Diophantine Equation(to which there's either an infinite number of solutions, or there are none), having no solutions means $$$X$$$ is not a multiple of $$$gcd(a,b)$$$.
So if $$$gcd(a,b)=1$$$ then there's no such X.
Else there will be many; infinitely many.
D can be solved using a trick: create a virtual vertex, let's denote it as n + 1, then for every vertex connect it to (n + 1)th vertex with an edge having a weight equal to the vextex's c value. Then the problem becomes finding the MST of a given graph.
You don't need a virtual vertex for Prim's algo.
I used Kruskal
Woah !! Awesome trick.
can you please tell how you deduced that mst will work here, I mean it looks like baccktracking with unioin find to me.
The competition was awesome !!
Maybe I'm the only one who solve B using dp and C using 2-dimension dp (I didn't think of Fibonacci or any stuffs related to it)
Do you use EternalFire ?
ofcourse it's a certificate I got last year after participating a programming contest and now it's my mousepad
In problem C, I was using square free Fibonacci series. But, I got WA on pretest 9. Can someone help me?? My solution
Square free Fibonacci series is wrong. You should use normal Fibonacci series.
Ok, I got my mistake. Thanks
This might overflow since both are
int:ans= (ans * k)%mod;I have used #define for that
Anyone notices that there seems to be no one (at least very very few people) that fails system test. Amazing contest!
Damn, these 4 ms help me become a candidate master!
I converted the problem C into "how many ways you can get a sum s using coins {1,2}"..then somehow didn't able to implement it in time...Its kind of sad
That is a very nice solution.
but I later figured it out that it will not work since,in counting number of ways a sum can be generated from fixed number of coins, order doesn't matters,but in our case order of selection will matter...eg. sum=3 can be generated from coin-sets {1,2} in two different ways {1,1,1} ans {1,2}(note that {1,2} and {2,1} are same) but the answer should be 3.It will convert to staircase problem counting number of ways to reach stair n if you can climb 1 or 2 stair at a time....correct me if I am wrong
Yes, I don't think you are wrong
Can someone help me with solution of D.. Here's my submission 64039124
My idea was to check if a node can be connected with any other node such that the cost for wiring is minimum and less than or equal to the cost to build a power station on that node. If such a node is available then construct an edge between them. Otherwise build a power station on that node.
Lastly, if a component has no such node which has power station in it , then build a power station that requires minimum cost in one of the node of the component..
It is generating way too much minimum cost for test case 13. I couldnt find why
Try this:
Ans is 15 and not 17
This solution also generates 15
This is the output
15
1
1
2
2 3
3 1
The correct answer is 18
Thanks :)
A very balanced contest with strong cases , thank you :)
Great round! Thank you)
I had following idea to D. Lets make minimum spanning tree of graph with edges $$$c(i, j) = (k_i+k_j) * d(i, j)$$$ for all $$$i , j$$$. Firstly lets add the cheapest plant. Then sort edges in non-increasing order and we will try to delete every edge. If we are considering edge $$$(x,y)$$$ let $$$c_1$$$ will be component with vert $$$x$$$ of tree without edge $$$(x, y)$$$ and $$$c_2$$$ will be component with vert $$$y$$$ of tree without edge $$$(x, y)$$$. Only one of this components has installed plant now. Let it be $$$c_2$$$ . Then find cost of the cheapest plant in $$$c_2$$$ and check if its better to add plant instead of edge from spanning tree. If its better to add plant, we delete edge $$$(x, y)$$$ and add plant.
But, my solution got WA13. I will be grateful if you point out my mistake.
Not entirely sure but could be that you are not using the "cost" of power plant while sorting, you are only using wiring cost while sorting? Both types of costs need to be consider.
Thanks DeliciousFlatChest !
That was a good contest !
However, she died when she was only in fifth grade so she is not smart enough for this.Wait, what codeforces?
:D
Maybe it's reference to this character.
Honestly, didn't read this part until you pointed out, XD...
Has anyone solved D using dynamic programming?
Another solution for D is the O(n^2) version of Prim's algorithm which works better on a dense graph. It doesn't need extra dummy nodes either. Is it the most efficient solution?
I solved D in this way after the round, but my implementation was slower than I expected.
Can anyone help me with problem D. I sumbited the same code twice, and in the constest y got TLE, but after resumbitting i got AC. https://codeforces.me/contest/1245/submission/64045463 https://codeforces.me/contest/1245/submission/64035229
Your AC code was just 4 ms away from TLE. It just got lucky. The execution time of the same code can differ by a few ms in different submissions. That is the reason behind TLE. Your code was not just lucky enough to get AC in contest time!
D is very much similar to this problem : https://vjudge.net/problem/LightOJ-1059
Although the contest was very well written and prepared, but still after seeing the next contest, I am like — "Finally, a div. 3 round...., Yay!"
Hello my solution to problem D is either bugged or incorrect. With bugs I can cope but I worry it might be simply incorrect. It fails to pass test 13 (wrong answer).
My idea:
Firstly, take each city and make network out of it, making total n networks. Assume that every network doesn't have electricity supplied to it.
I'm not worried about time complexity as above can be implemented in a fairly fast way. Just whether this should work or not.
Thanks for help in advance.
Hello :)
I didn't read your method, all I could do to help was stresstest, and I found a case where your code(I used the last submission) fails compared to my accepted solution.
4
25 11
23 11
31 11
30 25
41 43 44 48
3 3 4 8
145
3
1 3 4
1
1 2
143
2
1 4
2
1 2
1 3
Thanks!
Turns out my solution was bugged :)
How are you supposed to solve D using Kruskal's Algorithm?
When I sorted the 2,000,000 edges, I got TLE on Test 7. Is this only because I used Java, or is Kruskal too slow in general for this problem?
I've heard that inbuilt sorting function in java is sometimes slow(maybe it is a quick sort variant),i do not know otherwise...
I have a sort method that randomizes the numbers being sorted first, so that it is immune to the Quicksort TLE. I simply believe that the time constraints are a little too tough for Java, at least using Kruskal.
I used Prim's algorithm to pass the test case in just 202 MS though.
great job I love this round too
Can someone help me for problem B? I cannot find out where I am wrong? 64097713
The way you find the optimal sequence of moves(your string 'alice') in the case of YES is wrong.
Try this case to find what you did wrong.
1
3 2 1 0
SPS
YES
RPR