:(

1) кучи должны быть одинаковыми?
2) если нет то почему нельзя взять самое маленькое число и вычесть из него остальные?

А тестов ровно 128?

Only KS can be skipped but not KD KC KH

if you have a WA5, it is possible that you also did not read the condition, and instead of a 5000 bill you wrote something else (for example, 2000).

Edited by author 29.12.2024 14:51

Looks like Noise function acts like in real world. It is some random variable from normal distribution with zero expected value, and some small variation

Edited by author 27.12.2024 03:35

For a positive integer x, we define R(x) as a number equal to the inverted x in decimal notation. For example, R(123) = 321, R(1) = 1, and R(100500) = 5001.

For a given number S, calculate the number of integers x such that 1 ≤ x and x+R(x) ≤ S.

As an answer, print the remainder of dividing this number by 10^9 + 7.

Input data contains an integer S written without leading zeros (0 < S < 10^100000).

Test: 1000000 Answer: 505449

The idea: the dynamics of the digits of the number?
What other ideas can you suggest?

Im using BFS to find shortest way from S to F, BFS to count dists, BFS to count probability on shortest ways. I tried c++ vectors and arrays, both are getting TLe

Maybe cause is long double values (three BFS can pass in 1 second). How can I fix it? Maybe Dijkstra on two parameters?

O(N*logN) or O(N+max(P))?

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.