Hello fellow Codeforcers ! I took part in [contest:2033] on last Friday. I didn't have a Div. 3 contest in a long time, so I was eager to participate. I didn't like the round so much, because I felt it relied too much on implementation rather than problem solving. I was able to find the theoretical solution for A-F during the contest, but I couldn't implement successfully D-F in real time. Having a better understanding of what I missed could get me a huge improve in future contests.↵
↵
This is why I share here is my thoughts and solutions about the first 6 problems of the contest. Welcome to my humble green editorial !↵
↵
↵
## [problem:2033A]↵
↵
<spoiler summary="Observation">↵
The position of the dot after the $k$-th turn is $(-1)^k*k$.↵
↵
<spoiler summary="Proof">↵
Let's denote $S_k = \sum_{i=1}^{i=k}{(-1)^i*(2*i-1)}$ the position of the dot after the $k$-th turn. We can show by induction that $S_k = (-1)^k*k$. ↵
↵
- It is true for $k = 0$.↵
↵
- Suppose it true for a fixed $k$. $S_{k+1} = S_k + (-1)^{k+1}*(2*k+1) = (-1 )^k*k + (-1)^{k+1}*(2*k+1) = (-1)^{k+1}*(k+1)$↵
</spoiler>↵
↵
</spoiler>↵
↵
We want to find the parity of the last $k$ s.t. $|S_k|\leq n \implies k \leq n$, i.e. the parity of $n$. If $n\%2$, Kosuke plays the last turn, else it is Sakurako.↵
↵
#### Complexity↵
$\mathcal{O}(1)$ complexity↵
↵
#### What I've learnt↵
Not much, Div. 3 A problem, focus on pattern identification. ↵
↵
## [problem:2033B]↵
↵
Observation : When defining a square to increase one value on its diagonal, it's always better to consider the largest square possible increasing it.↵
↵
The formalism here is boring and I think it doesn't really help to understand the solution. At each operation, one can increase all values in the same "diagonal" of the matrix by 1 by choosing the largest possible squares for each "diagonal". The answer is to find the sum of the maximum over all matrix "diagonals".↵
↵
#### Complexity↵
You'll end up iterating over all values in a specific order.↵
$\mathcal{O}(n^2)$ complexity.↵
↵
#### What I've learnt↵
Not much, problem statement was misleading for me, and the only difficulty is the implementation.↵
↵
↵
## [problem:2033C]↵
↵
I read many unhappy comments on C, and during the contest, less people succeeded at it than D. I felt it wasn't a hard problem at all, which means I must be improving :D .↵
↵
Observation : The participation to the disturbance of each pair of $(i, n-i+1)$ can be decided only based on the previous pair $(i-1, n-(i-1)+1)$ ↵
↵
This leads to consider pairs iteratively, starting from the middle pair (or middle point) of the array. For the $i$-th pair $(i, n-i+1)$, regardless of the positioning of the $(i-1)$-th pair, you can always swap it or not to get the minimal disturbance amount.↵
↵
#### Complexity↵
Iterating over values in $a$ results in $\mathcal{O}(n)$ complexity overall.↵
↵
#### What I've learnt↵
This was an easy problem in my opinion, but nothing surprising for a Div. 3 C problem.↵
↵
↵
## [problem:2033D]↵
↵
Here comes my big frustration. During the contest, I found the statement misleading, since there may be many way to chose beautiful segments. Once I got my head wrapped around it, the idea for this problem stroke me as pretty simple.↵
↵
<spoiler summary="Observation : ">↵
(Very standard) The sum over a segment $(l,r)$ can be computed as $pref[r] - pref[l-1]$ where $pref$ is a prefix sum array.↵
</spoiler>↵
↵
For any two positions $i < j$, if $pref[i] == pref[j]$, then $(i+1,j)$ is a beautiful segment. To find the maximum number of such segments, one should consider greedily the shortest possible segments. This can be done by iterating of $i$ and store the prefixes met. If the current prefix has been seen after the last beautiful segment has been found, a new shortest non-overlapping beautiful segment can be considered.↵
↵
The loop is in $\mathcal{O}(n)$, and I used a $unordered\_set$ to insert and search elements in $\mathcal{O}(1)$.↵
↵
During the contest, I got WA (on 530-th value of test 2 ...) because I wasn't careful enough on handling beautiful segments with ends in $1$ or $n$. ↵
↵
I came back after the contest, and wrote correct code, this time getting TLE on test 18. My solution seems correct and I didn't know how to optimize it.↵
↵
<spoiler summary="Code">↵
~~~~~↵
#include <bits/stdc++.h>↵
#include <vector>↵
↵
↵
using namespace std;↵
typedef long long ll;↵
typedef unsigned long long ull;↵
↵
#define forn(i,j,k) for(ll i=(j); i<=(k); i++)↵
#define rofn(i,j,k) for(ll i=(j); i>=(k); i--)↵
#define forv(b,a) for(auto &b : a)↵
↵
int main() {↵
int t;↵
cin >> t;↵
↵
ll n;↵
while(t--) {↵
cin >> n;↵
vector<ll> a(n);↵
forn(i,0,n-1) cin >> a[i];↵
↵
↵
ll pref = 0;↵
unordered_set<ll> p;↵
ll ans = 0;↵
forn(i,0,n-1) {↵
pref += a[i];↵
if(p.find(pref) != p.end() || a[i] == 0) {↵
ans++;↵
p.clear();↵
} else {↵
p.insert(pref);↵
}↵
}↵
↵
cout << ans << endl;↵
}↵
↵
return 0;↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
↵
Looking at others submissions, I came across(TLE) and [this one](https://mirror.codeforces.com/contest/2033/submission/288070699)(Accepted) by [user:avi032904,2024-11-04], between which he changed from an unordered_set to a set. I did some more research and found [this blog](https://mirror.codeforces.com/blog/entry/62393) by [user:neal,2024-11-04]. I highly encourage you to read it, I'll summarize here the main ideas.↵
↵
For insertion operations on an unordered_map to be in $\mathcal{O}(1)$ time complexity, the average number of collisions for each item should be $\mathcal{O}(1)$ on average. This is most likely always true in an context where items at selected at random. Now, in a hacking context, knowing the hash function of the unordered_map, one can deliberately cause as much collisions as possible and make insertion a $\mathcal{O}(n^2)$ operation.↵
↵
An approach is to reintroduce unpredicability by using a precise clock and some arithmetic operations to prevent targeted attacks on modular key values. The detailed explanation is to read on neal's blog.↵
↵
This allows my code to pass all tests.↵
↵
<spoiler summary="My final code">↵
↵
~~~~~↵
#include <bits/stdc++.h>↵
#include <vector>↵
↵
↵
using namespace std;↵
typedef long long ll;↵
typedef unsigned long long ull;↵
↵
#define forn(i,j,k) for(ll i=(j); i<=(k); i++)↵
#define rofn(i,j,k) for(ll i=(j); i>=(k); i--)↵
#define forv(b,a) for(auto &b : a)↵
↵
struct custom_hash {↵
static uint64_t splitmix64(uint64_t x) {↵
// http://xorshift.di.unimi.it/splitmix64.c↵
x += 0x9e3779b97f4a7c15;↵
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;↵
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;↵
return x ^ (x >> 31);↵
}↵
↵
size_t operator()(uint64_t x) const {↵
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();↵
return splitmix64(x + FIXED_RANDOM);↵
}↵
};↵
↵
int main() {↵
int t;↵
ios_base::sync_with_stdio(false);↵
cin.tie(NULL); cout.tie(NULL);↵
cin >> t;↵
↵
ll n;↵
while(t--) {↵
cin >> n;↵
vector<ll> a(n+1);↵
forn(i,1,n) cin >> a[i];↵
↵
ll pref = 0;↵
unordered_map<ll, ll, custom_hash> m;↵
ll ans = 0;↵
ll last = -1;↵
forn(i,0,n) {↵
pref += a[i];↵
if(m.find(pref) != m.end() && m[pref] >= last) {↵
ans++;↵
last = i;↵
}↵
m[pref] = i;↵
}↵
↵
cout << ans << endl;↵
}↵
↵
return 0;↵
}↵
↵
~~~~~↵
↵
↵
</spoiler>↵
↵
#### Complexity↵
The code only performs one prefix sum on the whole array and perform basic operations on an unordered_map, resulting in an overall $\mathcal{O}(n)$ complexity.↵
↵
#### What I've learnt↵
In some (hacking) contexts, things I held to be always true (insertion in $\mathcal{O}(1)$) can outcome being proven wrong. When facing such situations, one must be ready to look into a new level of complexity and understand in depth the related concepts. I'll definitely remember this one.↵
↵
## [problem:2033E]↵
↵
This is not the first time I see a problem about permutations with an operation involving $p_{p_i}$. I should remember the following observations. My editorial is very similar to the official from [user:FBI,2024-11-04] :↵
↵
<spoiler summary="Observations">↵
Observation 1 : A \textit{simple} permutation is a permutation where all cycles are of length 1 or 2. ↵
↵
Observation 2 : When swapping two elements that belong to the same cycle, two smaller cycles are created. A consequence is that each cycle can be independently broken into smaller parts.↵
</spoiler>↵
↵
↵
My idea during the contest was to suppose that one was always able to break a cycle of size $k$ into two cycles of size $\frac{k}{2}$ and $\frac{k}{2}+(k\%2)$. I used a recursive function and memoization to get the statement tests right, but got WA on testcase 2.↵
↵
My assumption and my implementation were correct, but it doesn't yield the optimal answer. A better assumption is that is is always possible to perform a swap to create 2-cycles iteratively. ↵
↵
<spoiler summary="Proof :">↵
Consider a permutation $[p_1, p_2, p_3, ..., p_n]$ with a cycle of size $k \geq 3$ denoted by $(c_1, c_2, ..., c_k)$. This can be seen as the sequence of consecutive visited values, and it holds that $p_{c_i} = c_{i+1}$.↵
↵
If we swap the positions of the values $c_1$ and $c_3$ :↵
↵
- $p_{c_k}$ yields the value in positions $c_k$, which was $c_1$ and is now $ c_3$↵
↵
- $p_{c_2}$ yields the value in $c_2$ which was $c_3$ and is now $c_1$↵
↵
- $\forall i \neq 2,k, p_{c_i} = c_{i+1}$↵
↵
We then have a cycle $(c_2, c_3)$ of size $2$ and a cycle $(c_1,c_4,...,c_k)$ of size $k-2$↵
</spoiler>↵
↵
↵
A cycle of size $n$ needs $\lfloor \frac{k-1}{2} \rfloor$ operations to create a valid subsequence in the permutation.↵
↵
#### Complexity↵
Detecting all cycles is in $\mathcal{O}(n)$ and each can be handled independently, resulting in an overall $\mathcal{O}(n)$ complexity.↵
↵
#### What I've learnt↵
My first assumption was suboptimal, and I failed at realizing it in time. This lead me to prove the result used in the problem, I'll probably have use for this later.↵
↵
## [problem:2033F] ↵
↵
This problem is definitely built like a Project Euler problem, which is not to my displeasure. I took some 30 minutes during the contest to think about this, and I must say that I came pretty close. I submitted an accepted solution short after the end of the contest, which I already see as a win.↵
↵
I got the intuition that Fibonacci modulo $k$ was periodic. My approach was then to compute the first complete period and store the positions of zeros in an array $zeros$. The answer can be found by taking computing $\lfloor \frac{n}{size(z)} \rfloor$ to know the number of complete periods before $n$, and consider the right element in $zeros$.↵
↵
With more research and the official editorial, I learnt more properties about the Fibonacci sequence :↵
↵
- The Fibonacci modulo $n$ is called Pisano period and writes $\pi(n)$. ↵
↵
- $gcd(F_n, F_m) = F_{gcd(n,m)}$. This leads to $gcd(F_i, F_{i*j}) = F_i$ and finally $F_i | F_{i*j}$↵
↵
We can show that zeros in the period cycle are evenly spaced. $F(0) = 0$, we can skip it. If we denote $i$ the smallest natural integer such that $F(i) \equiv 0 [n]$, all others zeros will be at indices $i*j$.↵
↵
<spoiler summary="Proof">↵
- $F_i \equiv 0 [n]$, then $F_{i*j} \equiv 0 [n]$, so all positions $i*j$ are zeros.↵
↵
- Suppose we have $i$ and $j$ such that $F_i \equiv 0 [n]$, $F_j \equiv 0 [n]$ and $gcd(i,j) = 1$. Using the second property, $gcd(F_i, F_j) = \lambda * n = F_1$, which is impossible. In conclusion all zeros are evenly spaced.↵
↵
Finally, we proved that all zeros are evenly spaced by $i$ steps, with $i$ the index of the first $0$.↵
</spoiler>↵
↵
↵
The simplest solution is to look for $i$ and consider $i*n$.↵
↵
#### Complexity↵
It appears that $\pi(k) \leq 6*k$ the resulting complexity is then $\mathcal{O}(k)$↵
↵
#### What I've learnt↵
I really liked this type of problem. I have a computer science and mathematics background, so I should be able to be good at this type of problems. I will definitely try to remember these properties on Fibonacci (as well as the proof).↵
↵
↵
## Conclusion↵
↵
I have less time currently to write these custom editorials and I apologize for there is less preparation. I think it still allows me to better learn from contests and maybe share some thoughts with other participants. I really liked the problems and learnt many very useful things on maps, Fibonacci and others. Thank you again [user:FBI,2024-11-04], [user:Vladosiya,2024-11-04] and all other involved in the contest organization ([user:Parag_AP,2024-11-04] for green testing :)) and once again [user:MikeMirzayanov,2024-11-04] for creating Codeforces and Polygon.<br>↵
See you soon for my next editorial, this time with more graphics I promise.
↵
This is why I share here is my thoughts and solutions about the first 6 problems of the contest. Welcome to my humble green editorial !↵
↵
↵
## [problem:2033A]↵
↵
<spoiler summary="Observation">↵
The position of the dot after the $k$-th turn is $(-1)^k*k$.↵
↵
<spoiler summary="Proof">↵
Let's denote $S_k = \sum_{i=1}^{i=k}{(-1)^i*(2*i-1)}$ the position of the dot after the $k$-th turn. We can show by induction that $S_k = (-1)^k*k$. ↵
↵
- It is true for $k = 0$.↵
↵
- Suppose it true for a fixed $k$. $S_{k+1} = S_k + (-1)^{k+1}*(2*k+1) = (-1 )^k*k + (-1)^{k+1}*(2*k+1) = (-1)^{k+1}*(k+1)$↵
</spoiler>↵
↵
</spoiler>↵
↵
We want to find the parity of the last $k$ s.t. $|S_k|\leq n \implies k \leq n$, i.e. the parity of $n$. If $n\%2$, Kosuke plays the last turn, else it is Sakurako.↵
↵
#### Complexity↵
$\mathcal{O}(1)$ complexity↵
↵
#### What I've learnt↵
Not much, Div. 3 A problem, focus on pattern identification. ↵
↵
## [problem:2033B]↵
↵
Observation : When defining a square to increase one value on its diagonal, it's always better to consider the largest square possible increasing it.↵
↵
The formalism here is boring and I think it doesn't really help to understand the solution. At each operation, one can increase all values in the same "diagonal" of the matrix by 1 by choosing the largest possible squares for each "diagonal". The answer is to find the sum of the maximum over all matrix "diagonals".↵
↵
#### Complexity↵
You'll end up iterating over all values in a specific order.↵
$\mathcal{O}(n^2)$ complexity.↵
↵
#### What I've learnt↵
Not much, problem statement was misleading for me, and the only difficulty is the implementation.↵
↵
↵
## [problem:2033C]↵
↵
I read many unhappy comments on C, and during the contest, less people succeeded at it than D. I felt it wasn't a hard problem at all, which means I must be improving :D .↵
↵
Observation : The participation to the disturbance of each pair of $(i, n-i+1)$ can be decided only based on the previous pair $(i-1, n-(i-1)+1)$ ↵
↵
This leads to consider pairs iteratively, starting from the middle pair (or middle point) of the array. For the $i$-th pair $(i, n-i+1)$, regardless of the positioning of the $(i-1)$-th pair, you can always swap it or not to get the minimal disturbance amount.↵
↵
#### Complexity↵
Iterating over values in $a$ results in $\mathcal{O}(n)$ complexity overall.↵
↵
#### What I've learnt↵
This was an easy problem in my opinion, but nothing surprising for a Div. 3 C problem.↵
↵
↵
## [problem:2033D]↵
↵
Here comes my big frustration. During the contest, I found the statement misleading, since there may be many way to chose beautiful segments. Once I got my head wrapped around it, the idea for this problem stroke me as pretty simple.↵
↵
<spoiler summary="Observation
(Very standard) The sum over a segment $(l,r)$ can be computed as $pref[r] - pref[l-1]$ where $pref$ is a prefix sum array.↵
</spoiler>↵
↵
For any two positions $i < j$, if $pref[i] == pref[j]$, then $(i+1,j)$ is a beautiful segment. To find the maximum number of such segments, one should consider greedily the shortest possible segments. This can be done by iterating of $i$ and store the prefixes met. If the current prefix has been seen after the last beautiful segment has been found, a new shortest non-overlapping beautiful segment can be considered.↵
↵
The loop is in $\mathcal{O}(n)$, and I used a $unordered\_set$ to insert and search elements in $\mathcal{O}(1)$.↵
↵
During the contest, I got WA (on 530-th value of test 2 ...) because I wasn't careful enough on handling beautiful segments with ends in $1$ or $n$. ↵
↵
I came back after the contest, and wrote correct code, this time getting TLE on test 18. My solution seems correct and I didn't know how to optimize it.↵
↵
<spoiler summary="Code">↵
~~~~~↵
#include <bits/stdc++.h>↵
#include <vector>↵
↵
↵
using namespace std;↵
typedef long long ll;↵
typedef unsigned long long ull;↵
↵
#define forn(i,j,k) for(ll i=(j); i<=(k); i++)↵
#define rofn(i,j,k) for(ll i=(j); i>=(k); i--)↵
#define forv(b,a) for(auto &b : a)↵
↵
int main() {↵
int t;↵
cin >> t;↵
↵
ll n;↵
while(t--) {↵
cin >> n;↵
vector<ll> a(n);↵
forn(i,0,n-1) cin >> a[i];↵
↵
↵
ll pref = 0;↵
unordered_set<ll> p;↵
ll ans = 0;↵
forn(i,0,n-1) {↵
pref += a[i];↵
if(p.find(pref) != p.end() || a[i] == 0) {↵
ans++;↵
p.clear();↵
} else {↵
p.insert(pref);↵
}↵
}↵
↵
cout << ans << endl;↵
}↵
↵
return 0;↵
}↵
~~~~~↵
↵
</spoiler>↵
↵
↵
Looking at others submissions, I came across
↵
For insertion operations on an unordered_map to be in $\mathcal{O}(1)$ time complexity, the average number of collisions for each item should be $\mathcal{O}(1)$ on average. This is most likely always true in an context where items at selected at random. Now, in a hacking context, knowing the hash function of the unordered_map, one can deliberately cause as much collisions as possible and make insertion a $\mathcal{O}(n^2)$ operation.↵
↵
An approach is to reintroduce unpredicability by using a precise clock and some arithmetic operations to prevent targeted attacks on modular key values. The detailed explanation is to read on neal's blog.↵
↵
This allows my code to pass all tests.↵
↵
<spoiler summary="My final code">↵
↵
~~~~~↵
#include <bits/stdc++.h>↵
#include <vector>↵
↵
↵
using namespace std;↵
typedef long long ll;↵
typedef unsigned long long ull;↵
↵
#define forn(i,j,k) for(ll i=(j); i<=(k); i++)↵
#define rofn(i,j,k) for(ll i=(j); i>=(k); i--)↵
#define forv(b,a) for(auto &b : a)↵
↵
struct custom_hash {↵
static uint64_t splitmix64(uint64_t x) {↵
// http://xorshift.di.unimi.it/splitmix64.c↵
x += 0x9e3779b97f4a7c15;↵
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;↵
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;↵
return x ^ (x >> 31);↵
}↵
↵
size_t operator()(uint64_t x) const {↵
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();↵
return splitmix64(x + FIXED_RANDOM);↵
}↵
};↵
↵
int main() {↵
int t;↵
ios_base::sync_with_stdio(false);↵
cin.tie(NULL); cout.tie(NULL);↵
cin >> t;↵
↵
ll n;↵
while(t--) {↵
cin >> n;↵
vector<ll> a(n+1);↵
forn(i,1,n) cin >> a[i];↵
↵
ll pref = 0;↵
unordered_map<ll, ll, custom_hash> m;↵
ll ans = 0;↵
ll last = -1;↵
forn(i,0,n) {↵
pref += a[i];↵
if(m.find(pref) != m.end() && m[pref] >= last) {↵
ans++;↵
last = i;↵
}↵
m[pref] = i;↵
}↵
↵
cout << ans << endl;↵
}↵
↵
return 0;↵
}↵
↵
~~~~~↵
↵
↵
</spoiler>↵
↵
#### Complexity↵
The code only performs one prefix sum on the whole array and perform basic operations on an unordered_map, resulting in an overall $\mathcal{O}(n)$ complexity.↵
↵
#### What I've learnt↵
In some (hacking) contexts, things I held to be always true (insertion in $\mathcal{O}(1)$) can outcome being proven wrong. When facing such situations, one must be ready to look into a new level of complexity and understand in depth the related concepts. I'll definitely remember this one.↵
↵
## [problem:2033E]↵
↵
This is not the first time I see a problem about permutations with an operation involving $p_{p_i}$. I should remember the following observations. My editorial is very similar to the official from [user:FBI,2024-11-04] :↵
↵
<spoiler summary="Observations">↵
Observation 1 : A \textit{simple} permutation is a permutation where all cycles are of length 1 or 2. ↵
↵
Observation 2 : When swapping two elements that belong to the same cycle, two smaller cycles are created. A consequence is that each cycle can be independently broken into smaller parts.↵
</spoiler>↵
↵
↵
My idea during the contest was to suppose that one was always able to break a cycle of size $k$ into two cycles of size $\frac{k}{2}$ and $\frac{k}{2}+(k\%2)$. I used a recursive function and memoization to get the statement tests right, but got WA on testcase 2.↵
↵
My assumption and my implementation were correct, but it doesn't yield the optimal answer. A better assumption is that is is always possible to perform a swap to create 2-cycles iteratively. ↵
↵
<spoiler summary="Proof
Consider a permutation $[p_1, p_2, p_3, ..., p_n]$ with a cycle of size $k \geq 3$ denoted by $(c_1, c_2, ..., c_k)$. This can be seen as the sequence of consecutive visited values, and it holds that $p_{c_i} = c_{i+1}$.↵
↵
If we swap the positions of the values $c_1$ and $c_3$ :↵
↵
- $p_{c_k}$ yields the value in positions $c_k$, which was $c_1$ and is now $ c_3$↵
↵
- $p_{c_2}$ yields the value in $c_2$ which was $c_3$ and is now $c_1$↵
↵
- $\forall i \neq 2,k, p_{c_i} = c_{i+1}$↵
↵
We then have a cycle $(c_2, c_3)$ of size $2$ and a cycle $(c_1,c_4,...,c_k)$ of size $k-2$↵
</spoiler>↵
↵
↵
A cycle of size $n$ needs $\lfloor \frac{k-1}{2} \rfloor$ operations to create a valid subsequence in the permutation.↵
↵
#### Complexity↵
Detecting all cycles is in $\mathcal{O}(n)$ and each can be handled independently, resulting in an overall $\mathcal{O}(n)$ complexity.↵
↵
#### What I've learnt↵
My first assumption was suboptimal, and I failed at realizing it in time. This lead me to prove the result used in the problem, I'll probably have use for this later.↵
↵
## [problem:2033F] ↵
↵
This problem is definitely built like a Project Euler problem, which is not to my displeasure. I took some 30 minutes during the contest to think about this, and I must say that I came pretty close. I submitted an accepted solution short after the end of the contest, which I already see as a win.↵
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I got the intuition that Fibonacci modulo $k$ was periodic. My approach was then to compute the first complete period and store the positions of zeros in an array $zeros$. The answer can be found by taking computing $\lfloor \frac{n}{size(z)} \rfloor$ to know the number of complete periods before $n$, and consider the right element in $zeros$.↵
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With more research and the official editorial, I learnt more properties about the Fibonacci sequence :↵
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- The Fibonacci modulo $n$ is called Pisano period and writes $\pi(n)$. ↵
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- $gcd(F_n, F_m) = F_{gcd(n,m)}$. This leads to $gcd(F_i, F_{i*j}) = F_i$ and finally $F_i | F_{i*j}$↵
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We can show that zeros in the period cycle are evenly spaced. $F(0) = 0$, we can skip it. If we denote $i$ the smallest natural integer such that $F(i) \equiv 0 [n]$, all others zeros will be at indices $i*j$.↵
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<spoiler summary="Proof">↵
- $F_i \equiv 0 [n]$, then $F_{i*j} \equiv 0 [n]$, so all positions $i*j$ are zeros.↵
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- Suppose we have $i$ and $j$ such that $F_i \equiv 0 [n]$, $F_j \equiv 0 [n]$ and $gcd(i,j) = 1$. Using the second property, $gcd(F_i, F_j) = \lambda * n = F_1$, which is impossible. In conclusion all zeros are evenly spaced.↵
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Finally, we proved that all zeros are evenly spaced by $i$ steps, with $i$ the index of the first $0$.↵
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The simplest solution is to look for $i$ and consider $i*n$.↵
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#### Complexity↵
It appears that $\pi(k) \leq 6*k$ the resulting complexity is then $\mathcal{O}(k)$↵
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#### What I've learnt↵
I really liked this type of problem. I have a computer science and mathematics background, so I should be able to be good at this type of problems. I will definitely try to remember these properties on Fibonacci (as well as the proof).↵
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## Conclusion↵
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I have less time currently to write these custom editorials and I apologize for there is less preparation. I think it still allows me to better learn from contests and maybe share some thoughts with other participants. I really liked the problems and learnt many very useful things on maps, Fibonacci and others. Thank you again [user:FBI,2024-11-04], [user:Vladosiya,2024-11-04] and all other involved in the contest organization ([user:Parag_AP,2024-11-04] for green testing :)) and once again [user:MikeMirzayanov,2024-11-04] for creating Codeforces and Polygon.<br>↵
See you soon for my next editorial, this time with more graphics I promise.
