It seems, the only solution to this problem is recursion with correct "guess" on how to implement it to avoid TL.

Consider the fact that for any sequence $G$ with the form of fibonacci, that is: $G_{i+1} = G_i + G_{i-1}$, then $G_i = G_1 \cdot F_i + G_0 \cdot F_{i-1]$. So the problem became to solve two linear equations in integers. Prove it with induction or consider the matrix exponentiation for fibonacci.

You should use Python to avoid overflow (long long isn't sufficient, at least for my implementation). Good Luck!

This is painful problem, I spent 3 hours to solve, but very educational actually. If you are struggling, than read this: https://codeforces.com/topic/60789/ru9

"The vertices are listed in the order of traversal" means either in a clockwise or counterclockwise direction.

CODE IS HERE;
#include "stdio.h"
#include "string.h"
long a[401][401];
long n,h,t;
long i,j,m;
int  c[500];
int  p[500];

long count=0;
void bfs(int i)
{
    int k;
    for(k=1;k<=n;k++)
    {
        if(c[k]==0)
        {
            if(a[i][k]==1)
              {
                a[i][k]=0;
                a[k][i]=0;
                c[k]=1;
                p[k]=i;
                bfs(k);
               }
        }
    }

}


void path(int a1,int a2)
{
    int q[500];
    int i=0;
    while(1)
    {
        q[i]=a2;
        i++;
        a2=p[a2];
        if(a2==a1) break;
    }
    printf("%d",i+1);
    printf(" %d",a1);
    while(i>0)
    {
        printf(" %d",q[i-1]);
        i--;
    }
    printf("\n");


}



void main()
{
    for(i=0;i<401;i++)
        for(j=0;j<401;j++)
            a[i][j]=0;
    scanf("%ld %ld",&n,&m);
    for(i=0;i<m;i++)
    {
        scanf("%ld %ld",&h,&t);
            a[h][t]=1;
            a[t][h]=1;
    }
    for(i=1;i<=n;i++)
        c[i]=0;

    c[1]=1;
    bfs(1);

    for(i=1;i<n;i++)
    {
        for(j=i+1;j<=n;j++)
            if(a[i][j]==1)
            {
                count++;
            }
    }
    printf("%ld\n",count);
    for(i=1;i<n;i++)
    {
        for(j=i+1;j<=n;j++)
            if(a[i][j]==1)
            {
                path(i,j);
            }
    }

}

At the moment my way is to look through all the squares of length LEN (using binary search) and to check, whether a hash of a square was met before. It is of O(N*M*log(min{N, M})) complexity.
But I get WA#10 again and again. Please, if someone had a problem with this test, share your impressions, 'cause I'm going slightly mad.

got WA on 3, why, have any tests?

Edited by author 14.03.2024 17:39

easy max flow

5 plus 1 plus 1 plus 1 plus 1 plus 1 plus 3 plus 1 plus 1 plus 1 plus 1 plus 1 plus 1 is ...
19!!!!! NINETEEN!!!1111!! FFFFUUUUU~~~~~

Okay, have to read prob statement again. :D

Прошу уточнить.
В задании (и тестах) идёт речь о:
1) конечных географических 2D картах.
2) географических картах на сфере.
3) любых абстрактных картах в многомерном \ многосвязном пространстве.
4) другое.

Please clarify.
The task (and tests) are about:
1) Finite 2D geographical maps.
2) geographical maps on the sphere.
3) favorite abstract maps in a multi-dimensional \ multi-connected space.
4) other.

Admins, please look at this issue.
I suppose it is mistake (or bug in rating system), that I became 1st place in rating solved same number of tasks as ACRush, but did it only in 2024 year, i.e. later. Suppose that rating must be the same and my place is 2nd.

test:
1
10000
0

How do I answer "YES" or "NO"?

ooopsss! Zero not allowed! Sorry.

Edited by author 29.11.2024 01:34

когда я с длинной арифметикой сделал эту задачу я использовал в качестве основы 10000 и у меня был WA#7 тест
потом я сделал снову 10  и у меня был WA#2 тест
тогда подумав решил использовать EXTENDED всеравно WA#6 тест
кто подскажет как преодолеть 2, 6, 7 тесты

If you have an absolutely useless piece of code in your code that is not used anywhere, but it has an error going out of bounds, then this may instead affect the execution of the rest program

Round(time) instead of Ceil(time)

What should i check if i have WA in test 5?

Can't debug it, WA4/6

Delete a rectangle if there is another one covering this one

All the test cases I've come up with have been solved correctly by this method but no AC :(

code:
```
#include <bits/stdc++.h>

using namespace std;

int main() {
  int n;
  cin >> n;

  vector<vector<int>> x(n, vector<int>(3, 0));

  for (int i = 0; i < n; i++) {
    cin >> x[i][0];
    cin >> x[i][1];
    x[i][2] = i + 1;
  }

  sort(x.begin(), x.end());

  auto med = (n - 1) / 2;

  cout << x[med][2] << " " << x[med + 1][2];
}

```

Edited by author 11.11.2024 16:23

Edited by author 11.11.2024 16:23