GlebsHP's blog

By GlebsHP, history, 15 months ago,
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 » 15 months ago, # | ← Rev. 4 →   0 Very fast editorial! Thanks!
 » 15 months ago, # |   +105 I was too scared that $O(2^n \cdot n^2)$ is too slow for E so I ended up figuring it out in $O(2^n \cdot n)$ with $O(2^n)$ memory.
•  » » 15 months ago, # ^ |   +26 Could you explain this solution?
•  » » » 15 months ago, # ^ | ← Rev. 2 →   +78 For each subset $msk$, compute $dp[msk]$ as the possible last vertices $v$ of simple chains starting at $u$ and ending at $v$ (where $u$ is the minimum vertex in $msk$). Use bitwise operations to check for a given $msk$ all the new vertices $v'$ that can be added (condition is $graph[v'] \cap dp[msk] \neq \emptyset$).
 » 15 months ago, # |   +49 Can't solve D :(How to improve solving greed problems
•  » » 15 months ago, # ^ |   -26 It's not a greed problem!
•  » » » 15 months ago, # ^ |   +18 It is. At least my solution is
•  » » » » 15 months ago, # ^ |   +24 Hmm, it looks like it is. I had easier solving it with DP lol.
•  » » » » » 15 months ago, # ^ |   +5 how was ur dp
•  » » » » » 15 months ago, # ^ |   +3 Can you please explain your DP solution?
•  » » » » » » 15 months ago, # ^ |   -55 Nah
•  » » » » » » 15 months ago, # ^ |   0 Maximizing $T_2$ is easy, so the only problem is to minimize it. $T_0 + T_1 + T_2 = \frac{m}{4}$, so instead of minimizing $T_2$ you can maximize $T_0 + T_1$. You can do it by calculating $dp_i = \max T_0 + T_1$ on prefix $i$. Transitions are easy, I'm sure you can figure them out.
•  » » » » » 15 months ago, # ^ |   +38 Your solution is essentially greedy without you noticing it
•  » » » » » » 15 months ago, # ^ |   0 During the contest, I only realized that I need to maximize certain 2-sized flats without realizing that taking the one with minimal index always works.
•  » » 15 months ago, # ^ |   0 tags: greedy Ratings: 1700+ Should work
 » 15 months ago, # |   +13 Damn! Editorials are out even before the system test. Coolest round ever!(for me)
 » 15 months ago, # | ← Rev. 5 →   -6 Why my approach to problem C is wrong?We need for every $1 \le f \le p$ to check if: \begin{aligned} x + \sum_{i = 1}^{f} i &\equiv 0 \mod n \\ x + \frac{f (f + 1)}2 &\equiv 0 \mod n \\ f ^ 2 + f &\equiv -2x \mod n \end{aligned}We end up checking the value of $f ^ 2 + f$ (mod $n$), which is equivalent to checking $m ^ 2 + m$ whereas $m \equiv f \mod n$.From the previous reasoning, it is enough to check all possible values, $1 \le f \le \min(p, n)$.The vague part is going from $f$ to $m$. Why isn't it correct?
•  » » 15 months ago, # ^ | ← Rev. 2 →   +18 You can't multiply by 2 so you have to go up to min(p, 2n) not min(p, n)
•  » » » 15 months ago, # ^ | ← Rev. 2 →   +8 To be accurate, if $n$ is odd, $2$ is invertible $\mod n$ so you can multiply by $2$ and check only up to $n$ when it is odd.
•  » » » 15 months ago, # ^ |   0 can you explain why exactly I can't multiply by 2?
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15 months ago, # ^ |
Rev. 4   +19

This is because you want "$\iff$" (**if and only if**). Therefore, you need to be abble to multiply by $2$ (always works) but you also need to divide by $2$ (only works if $n$ is odd).

1) Multiplication by $2$ (always works)

Let $f \in \mathbb{Z}$ (an integer) and assume that $x + \dfrac{f(f + 1)}{2} = 0 \mod n$.

This means that $x + \dfrac{f(f + 1)}{2} = \lambda n$ for some $\lambda \in \mathbb{Z}$.

Therefore $2x + f(f + 1) = 2 \lambda n$.

Since $2 \lambda \in \mathbb{Z}$, $2x + f(f + 1) = 0 \mod n$.

2) Division by $2$ (only works if $n$ is odd)

Now, let's assume that $2x + f(f + 1) = 0 \mod n$. Can we say that $x + \dfrac{f(f + 1)}{2} = 0 \mod n$?

2.a) Assume $n$ is odd

Then $\dfrac{n + 1}{2}$ is an integer.

Thus, $\dfrac{n + 1}{2} (2x + f(f + 1)) = 0 \mod n$.

By simplifying, and since $\dfrac{f(f + 1)}{2}$ is an integer, you do obtain $x + \dfrac{f(f + 1)}{2} = 0 \mod n$.

2.b) Assume $n$ is even

Then $\dfrac{n + 1}{2}$ is not an integer so the above trick does not work.

What we would like to do is to deduce that $y = 0 \mod n$ from the fact that $2 y = 0 \mod n$.

However, remember that $n$ is even so $n = 2m$ for some $m$ in $\mathbb{Z}$. Taking $y = m$, $2 y = 0 \mod n$ but $y \neq 0 \mod n$.

Therefore, you cannot assume that $x + \dfrac{f(f + 1)}{2} = 0 \mod n$ if you only know that $2x + f(f + 1) = 0 \mod n$.

EDIT: I missclicked on post instead of preview, I'm going to finish writing my message. xD EDIT 2: I finished my message.

•  » » 15 months ago, # ^ | ← Rev. 2 →   0 I believe you have to check f up to min(p, 2n). I got the same equation as you, but I found out 2n is needed after trying out a few examples. I'd be happy to learn why 2n is needed. EDIT: See Beaboss's comment.
 » 15 months ago, # |   +12 You're given 3 problems and two hours to solve them
•  » » 15 months ago, # ^ | ← Rev. 2 →   0 D was not that difficult though
 » 15 months ago, # |   +8 I have trouble understanding the additional question at the end of the tutorial for problem c:"Question, can you build the test that required Vesper to use $x$ more than $100k$?"Can somebody explain the question please? (not the solution though, I wanna solve it ^^')
•  » » 15 months ago, # ^ |   0 I guess it’s asking you to find a case (i.e. find a $n$) for which only a $x > 10^5$ could make the answer “Yes” for an enough large $p$.Formally, find $(n, x, p)$ ($x > 10^5$) which would result “Yes” while would result “No” for any $x \le 10^5$.
•  » » » 15 months ago, # ^ |   0 But x
•  » » » » 15 months ago, # ^ |   0 Well, then it might be $f$ instead of $x$.Find (n, x) that requires a f $> 10^5$ to make (x + f(f+1)/2) % n == 0
•  » » » » » 15 months ago, # ^ |   0 Yes you got the right interpretation, thanks.There is indeed only one tuple $x, n$ with $0 \leqslant x \leqslant n - 1$ and $n \leqslant 100,000$ such that you can indeed reach $0$ starting from $x$ for some $f > 100,000$ such that you cannot do it for any $f \leqslant 100,000$.
 » 15 months ago, # |   0 why it will be tle ? i only use upper_bound get AC 197098569
•  » » 15 months ago, # ^ | ← Rev. 2 →   0 Delete the vector re from your code. Its size is big and it gets initialized for each test case.
 » 15 months ago, # |   0 it is logically very obvious that  (2n*(2n+1)/2)%n=0 So, logically it is obvious that if we try out values from 1 to 2*n and do not get answer then we would not get answer after that too. But can anyone prove it mathematically though? Still cannot convince myself about the 2*n thing.
•  » » 15 months ago, # ^ |   +10 The sum up to $f$ is the same as the sum up to $f + 2n$ ($\mod n$). $\dfrac{(f + 2n)(f + 2n + 1)}{2} = \dfrac{f (f + 1)}{2} + f n + n (f + 1) + n (2n + 1)$
•  » » » 15 months ago, # ^ |   0 do we try adding 2n to f by just intuition or their is something behind that?
•  » » » » 15 months ago, # ^ |   0 There is indeed something behing it (although the fact that $2n$ is going to work feels intuitive as well).By going back to the definition of $x = y \mod n$ (that is $x = y \mod n \iff \exists \lambda \in \mathbb{Z}, x = y + \lambda n$), you can prove that $x$ is a solution to $\dfrac{x}{a} = b \mod n$ if, and only if, $x$ is a solution to $x = ab \mod an$ (assuming that $x$ can be divided by $a$).
•  » » 15 months ago, # ^ | ← Rev. 5 →   0 Because $x + i(i+1)/2 \equiv x + (i+2n)(i+2n+1)/2$ ($mod$ $n$), the value of $x + i(i+1)/2$ $mod$ $n$ will repeat itself after $2n,$ so we only need to check from $1$ to $2n$ which is good enough
•  » » » 15 months ago, # ^ |   0 but why it didn't give tle when p reaches 2*n under 10^4 testcases as n=10^5
•  » » » » 15 months ago, # ^ |   +12 Because the sum of $n$ cannot be bigger than $2*10^5$
 » 15 months ago, # |   -18
 » 15 months ago, # |   +10 Easiest div2 d I've ever seen, I wish I registered for the contest
 » 15 months ago, # | ← Rev. 2 →   0 I think in task F you have used about 150 cpu-days
 » 15 months ago, # |   0 Why is this giving the wrong answer for the maximum value in problem D?
•  » » 15 months ago, # ^ |   +4 11110011?
•  » » » 15 months ago, # ^ |   0 I tried to improve further a bit. Can you point out my mistake? Here
•  » » » » 15 months ago, # ^ |   0 Try 1 12 111111100010 ans 5 8
•  » » 15 months ago, # ^ |   0 Would anyone be able to help me fix my code? I got the wrong answer on test 4 in this submission. I tried to generate all tests with n = 4 and n = 8, but I couldn't find why it was wrong
•  » » » 15 months ago, # ^ |   0 1 8 00111111 
 » 15 months ago, # |   0 :( Came home 1 hour late so couldn't join. Sad thing is I knew how to solve F because I did it in an approximation algorithms course before.
 » 15 months ago, # |   +159 F actually highlights something that is missing from Polygon (or maybe some online judges in general): the ability to declare that "this is the correct output file" without having it be generated by the model solution. In contrast, CMS (e.g. at IOI) treats the output files as the true ones and doesn't care if you didn't generate them using the model solution.This is a niche usage of course, but it would be nice to give approximation problems without hacks like "the last line is an encoded version of the true answer". Approximation algorithms aren't really the CP meta right now, but e.g. for one task in an ICPC-style contest, why not? Compute the correct answers using some slow brute force or a commercial ILP/SAT solver and the checker can just check the approximation factor.
•  » » 15 months ago, # ^ |   0 You can implement an approximation problem on Polygon using an interactor instead of exposing an encoded answer. The test case file contains the actual input to the problem and also the real answer. Interactor reads the test case file and only shows the actual input to the program. Then it compares the program's answer to the test case's answer. For this specific problem F it wouldn't allow for hacks because it's hard to implement a validator that checks we have the exact answer. However this method would allow hacks for many approximation problems such as minimum vertex cover.The problem could be "Given a graph, find a vertex cover that is at most twice the size of the minimum vertex cover." In this case we don't need a strict requirement that we have the optimal vertex cover in the test case, just some vertex cover. Then someone would be able to hack an incorrect program as long as they can find a vertex cover that is less than half the size of the program's answer.
 » 15 months ago, # | ← Rev. 4 →   -6 Randomised solution for E:DFS going deep for 2 random child every time. If you get loop — check if union of neighbours in it covers everything. If it does, then we found required loop in editorial. Repeat until you have spent almost 2 seconds. Print "No" otherwise. Not sure how to estimate probabilty of failure though.
 » 15 months ago, # |   +32 Btw, fun fact about problem F.There is a paper that proves that under SETH it is impossible to solve problem F in $O(m^{1-\varepsilon})$ time per query if we would need to answer queries online, there were also edge deletions, and we had to provide a $2-\delta$ approximation (here $f$-factor approximation means that we should provide an answer that is not smaller than the actual answer and at most $f$ times larger; I am not sure whether this theorem can be generalized to the notion of approximation that is used in problem F).
 » 15 months ago, # |   +3 In problem D it should be $T2=max(0,m4−∑yi=0⌊l′i2⌋)$ not $T2=min(0,m4−∑yi=0⌊l′i2⌋)$
 » 15 months ago, # |   0 The system test of E is weak. My brute force solution (with special check for cut-edges) passed it.
 » 15 months ago, # |   +41 By the way, there is another solution for F. Instead of running a binary search, you can just start bfs with updated vertexes and existing distances, not from scratch. This is too slow, but we can not run bfs each time, but only when sum of immediate increases (differences between end and start of added edges -1) is half of the current answer. This is correct because this sum is the upper bound of the answer decrease. The more tricky question why it is fast. After distance is less than $O(\sqrt(n))$, it is $O(nsqrt(n))$ even if we run bfs each time While difference is big, each edge is decreasing distance by at least factor 3/4 for at least $O(\sqrt(n))$ vertices. This looks like very rough bound, probably in practice it's better.
•  » » 15 months ago, # ^ |   0 Could you please provide more detail as to why 2. is true?
•  » » » 15 months ago, # ^ | ← Rev. 2 →   0 Hm. I was quite sure about it during the contest, but not really sure now. But I think it's still subquadratic. Idea is following. If we have a big shortcut (order of answer), then at least for second half of the vertices on the old shortest past to further end of shortcut, distance decreased significantly. Now we can say, that we either processed a lot of queries, or at least one of them is big. It seams it can lead to $N^{7/4}$ or $N^{5/3}$ bound. I'm not sure.
•  » » 15 months ago, # ^ |   0 Hi, I am a bit confused by what you call "sum of immediate increases".For instance, let's say I start the bfs from vertex $A$ and et's say I have two vertices $B$ and $C$ such $dist(A,B) = 42$ and $dist(A,C) = 50$. If I add the edge $(B, C)$, would the "immediate increase" be $50 - 42 = 8$?Also, when you say "this sum is the upper bound of the answer decrease", do you mean that the sum is an upper bound of the decrease of the diameter of the graph when adding the edge $(B, C)$?
 » 15 months ago, # |   0 what is the specific test in C? (last sentence in editorial)
•  » » 15 months ago, # ^ | ← Rev. 2 →   +3 65536 0 131071Did not expect to see a power of 2 tbh. Seems all n = 2^k with x = 0 will need 2*n - 1
 » 15 months ago, # |   +3 Problem C has a brilliant name. No thinking, just pull your luck ))
 » 15 months ago, # | ← Rev. 2 →   0 In Problem C, I was shocked. I write AC code but I thought it was WA. Because I thought the loop needed p time every test case. if I loop run P time then I got TLE. That's why I didn't submit my code. but after finishing the contest I submit my code and it was an AC code. void solve() { ll n, x, p; cin >> n >> x >> p; ll rem = n - x, sum = 0; if( x == 0 ) rem = 0; vector< int > v; for( ll i = 1 ; i <= n * 2 ; i++ ) { sum += i; if( sum % n == rem ) { cout << "YES" << endl; return; } if( i == p) break; } cout << "NO" << endl; } 
•  » » 15 months ago, # ^ | ← Rev. 2 →   0 because of this It is guaranteed that the sum of n over all test cases doesn't exceed 2⋅10^5
 » 15 months ago, # |   +48 Bonus question, can you guess the total number of cpu-days we have used to compute all the answers? Post you version in the comments! To find the diameter of a general graph, we will need $n$ BFS, and each takes $O(n+m)$ time, overall $O(n*(n+m))$ per query. $O(n*q*(n+m+q))$ per testcase.This problem has 80 testcases, assuming worst case $n=m=q=10^5$, its $80*10^5*10^5*(3*10^5) = 2.4*10^{17}$, assuming $10^8$ operations per second, on the normal computer we need $2.4*10^9$ seconds = ~3e5 days. I was expecting the answer for generating testcases to be < 100s, using 1e5 parallel cpus.
•  » » 15 months ago, # ^ |   +206 Both authors retired from CF in 2017, guess that's because they have been busy generating the test cases ever since...
•  » » » 15 months ago, # ^ |   +21
•  » » 15 months ago, # ^ | ← Rev. 2 →   +10 I think there are better algorithms / heuristics to enormously speed up the brute force. For example: If you add an edge, for the current diameter, you run BFS from one of the endpoints and if the edge addition doesn't improve the diameter, you don't need to calculate anything at this step of the algorithm anymore. Also, things like keeping some upperbounds of diameters ending in nodes, and not running a BFS if isn't strictly needed. Maybe not worstcase complexity improvements but they will speed up the brute force a whole lot.
•  » » » 15 months ago, # ^ | ← Rev. 6 →   +36 I have a solution that works in $O(nm + q(n + m + q))$ for all tests except special ones.The idea is very simple, at the very beginning we calculate the furthest distance from each vertex with $n$ BFSs and call this value upper bound.Now, after adding the next edge, we go through the vertices in descending order of upper bound and recalculate the furthest vertex from them. If it has not changed, this upper bound is the answer after adding edge.But unfortunately, it's still not fast enough and an additional hack is needed. And this hack is simple — in addition to recalculating the farthest vertex, let's update the upper bound of other vertices $u$ as "max distance from current vertex" + "distance from the current vertex to $u$". This is already enough to pass all the tests except for special ones, which are essentially built against such solution (at the moment, my solution can generate an answer to every test except two special in an hour).But it was not used as a model solution, but was rather a validator for validating the simplest distributed algorithm. Сode snippet for better understandingvoid init() { cerr << fixed << setprecision(5); start = clock(); layer_id = n; for (int i = 0; i < n; ++i) { if (i % 1000 == 0) { debug("init ", i); } ans[i] = naive::bfs(i); ans_per_layer[ans[i].dist].insert(ans[i].start); } } void update_ans(int start, int dist) { if (dist >= ans[start].dist) { return; } ans_per_layer[ans[start].dist].erase(ans[start].start); ans[start].dist = dist; ans_per_layer[ans[start].dist].insert(ans[start].start); } void recalc_ans(int v) { int mx = naive::bfs(v); for (int i = 0; i < n; ++i) { if (i != v) { update_ans(i, mx + naive::dst[i]); } else { update_ans(i, mx); } } } int solve(int hint = -1) { if (hint != -1) { recalc_ans(hint); } while (true) { while (ans_per_layer[layer_id].empty()) { --layer_id; } int cur_v = *ans_per_layer[layer_id].begin(); recalc_ans(cur_v); if (ans[cur_v].dist == layer_id) { break; } } return layer_id; } 
 » 15 months ago, # |   +34 Bonus question, can you guess the total number of cpu-days we have used to compute all the answers? Post you version in the comments! 7
 » 15 months ago, # |   0 How to solve A using DFS?
•  » » 15 months ago, # ^ |   0 it might time out, but you can keep the track of the path you have traveled, once you reach the destination you can compute the cost of that path.
 » 15 months ago, # | ← Rev. 2 →   0 Is there a ")" missing here for D?and isn't max?
•  » » 15 months ago, # ^ |   0 yes its max in t2
 » 15 months ago, # |   0 thank you for fast editorial
 » 15 months ago, # |   0 Why my solution got TLE even if it is O(nlogn).My Solution: https://mirror.codeforces.com/contest/1804/submission/197105822 Can someone help me, please?
•  » » 15 months ago, # ^ |   0 Hint: There's one line wrong, causing you to get stuck in an infinite loop SpoilerHint: It's line 43 SpoilerIt should be i+=k-1;
 » 15 months ago, # |   0 can someone explain why we are counting "at least one dark window segment" in problem D rather than one dark window segment .
 » 15 months ago, # |   0 In D, in last line, I think it should be max not min.
 » 15 months ago, # |   +42 F is a beautiful problem with a very scary statement
 » 15 months ago, # |   +8 cool way to get the correct answers in cool problem F
 » 15 months ago, # |   +55 Problem E: I was blown by the idea of using the first set bit to represent start of the circle. Very clever.
 » 15 months ago, # | ← Rev. 2 →   0 In problem D, my approach for finding the maximum value is to, first calculate "01" or "10", then "00" and at last "11", and assign this as two widowed rooms, and the remaining ones are single windowed. Can anyone point out my mistake? submission
 » 15 months ago, # |   0 What I do in C is, I run a loop from i = 1 to 1e5 and checking if i*n-x exist as sum of p natural number in logP overall time complexity is O(n*logP) but still it got TLE Why?https://mirror.codeforces.com/contest/1804/submission/197293170
•  » » 15 months ago, # ^ |   0 Note that you only loop from 1 to 1e4,so it would goes wrong when n is large enough.If the answer is No,it will run the loop completely,and your function f is $O(\log p)$,so the total is $t\times 10^4\times \log p=3\times 10^9$ and it goes tle.At least you should replace 1e4 with n*2.BTW,I'm not sure whether it's right or not to use binary search.
 » 15 months ago, # |   +5 I have received a mail that my solution of C problem in Nebius Welcome Round is coinciding with Somebody else . In mail it was written , to post about the evidence .C Problem was not too difficult for a person with average coding skills . The only trick in C was to get a control over the loop . I have seen many tutorials of past contests and in C problem , I used that for controlling the loop as the most common approach in these types get solved in nlog(n) or n root(n) . So I did it for n root(n) and then the code was simple a Brute force logic . It took me a number of contest and practice to reach specialist . Please reply me if i need to submit any supporting proof and what else can i upload . Please help GlebsHP , MikeMirzayanov . please do reply.
 » 15 months ago, # |   0 For Problem C.. I want to understand the repetitive behavior of the function (sum_from_1_to_2n % n), Please mention a mathematical or intuitive (better :) reasoning..
•  » » 15 months ago, # ^ |   0 Read my comment: https://mirror.codeforces.com/blog/entry/113830?#comment-1012138
 » 15 months ago, # |   +5 F is really fascinating.
 » 15 months ago, # |   +8 Can anyone give me any hint for problem H? Is the problem approachable with bitmask dp and meet in the middle?
•  » » 15 months ago, # ^ | ← Rev. 2 →   0 H can be solved with bitmask DP, but you have to come up with some methods to sum up the costs cleverly.
•  » » » 15 months ago, # ^ |   0 The problem I am facing is, I have two option, counter-clockwise and clockwise. I mean if its counter clockwise i have to add (a — b) * cnt[i][j] and (b — a) * cnt[i][j] if its clockwise. If i had only clockwise/ counter clockwise, i could have split the equation and take the sum independently (a * cnt[i][j] — b * cnt[i][j] for only counter clockwise). But now, how do i know which one is clock wise and which one is counter clockwise.
•  » » » » 15 months ago, # ^ |   0 That's the main difficulty of the problem. An approach is to consider the problem not from the perspective of characters, but from the edges on the cycle, that is, how many times each of them is used.
 » 15 months ago, # |   0 There is a formula mistake in tutorial of problem F. Having the correct value of $c_(G_i)(v)$ we can use the binary search ... That should be: Having the correct value of $c_{G_i}(v)$ we can use the binary search ...
•  » » 15 months ago, # ^ |   0 Fixed, thx
 » 15 months ago, # |   0 Auto comment: topic has been updated by GlebsHP (previous revision, new revision, compare).
•  » » 15 months ago, # ^ |   0 Tutorial is not available Why:(
•  » » » 15 months ago, # ^ |   +1 Looks like it takes some minutes for Codeforces to load the editorial from polygon's package. It is available now
 » 15 months ago, # |   0 Simple DP for D... why is the editorial so complicated?
•  » » 15 months ago, # ^ |   0 Could you explain your DP approach?
 » 15 months ago, # |   0 Auto comment: topic has been updated by GlebsHP (previous revision, new revision, compare).
 » 15 months ago, # | ← Rev. 2 →   0 I am very curious how the data of problem F was made, LOLIt is a really interesting problem, lots of thanks to the problem provider!
 » 7 months ago, # | ← Rev. 2 →   0 For the editorial of problem C, can you prove the following line? "The remainders of x of modulo n will repeat after 2n steps". In other words, How can you prove that, sums of all integers from 1 to i (1 <= i <= 2*n) will generate all the remainders from 0 to n-1, when diving by n?
•  » » 6 months ago, # ^ |   0 Check out my comment, you will understand it easily!
 » 6 months ago, # |   0 Question, can you build the test that required Vesper to use x more than 100k ? There is exactly one such test.This is from the tutorial of solution Of C(PULL YOUR LUCK) Could someone please help me to understand that case
 » 6 months ago, # | ← Rev. 3 →   0 Great problem C! Would like to share my approach. I observed a fairly obvious thing, for f=1, you will land on (x+1)%n [obviously], after that you can keep adding i (2<=i<=n) to the latest number. Eg: For n=5, x=2: 2 (+1)-> 3 (+2)-> 0 (+3)-> 3 (+4)-> 2 (+5)-> 2 , so now you again reached your starting point (2 in this case). This will hold for any odd n, as n*(n+1)/2 will by k*n which is congruent to 0 (mod n). So after n such steps you will always reach the starting x. After that the pattern pretty much loops, so we don't need to add more elements. For even n, this will not be true! (Reason: n*(n+1)/2 and n=2k => k*(2k+1) which need not give remainder 0 with 2k), but what will hold now is the same pattern twice! For eg: n=4, x=2: 2 (+1)-> 3 (+2)->1 (+3)->0 (+4)->0 (+1)->1 (+2)->3 (+3)->2 (+4)->2 !! (see how after one chain we reached 0 and not 2, but after 2 such operations we reached 2 again, proof is pretty simple, now as n=2k => 2n = 4k, so 4k*(4k+1)/2 = 2k*(4k+1) which is congruent to 0 (mod n=2k).After that you can store the n values (for odd case) and 2n values (for even case), and find the minimum index of 0. If there is no 0, that means for any p, you cant ever win. Else if p>= min_index : yes Else: no Here is my code for reference. Hope this explains well!