Maybe you forgot that the order of the numbers in the answer is important.

you can implement your simple fraction class and use it instead of floating point numbers

I don't know why, but in Python, code points of frame characters are:
upper_left  = 1066
upper_right = 1111
lower_left  = 1040
lower_right = 1065
horizontal  = 1044
vertical    = 1110
That's really strange, because there are no common codecs encode those characters as such.
Could Admin fix this or at lease give some explanations please?

Also, don't put non-ascii characters in your source code, but use:
  c == chr(1066)
or
  ord(c) == 1066

Any submission on Kotlin or Java return 'Restricted function' error. Even solutions that previously could pass at least some tests.

Здравствуйте, решение из примеров для задачи "1000. A+B"

object Main extends App {
   println(readLine().split(" ").map(_.toInt).sum)
}

выдает ошибку "Restricted function"

Edited by author 23.06.2021 07:43

Edited by author 23.06.2021 07:43

Edited by author 23.06.2021 07:43

Remember: maximum accuracy can not be more than 10 in minus the secong degree.
Запомните: максимальная точность числа не больше 10 в -2 степени.

I have O(len(s) * k^2), and it is definitely too slow (TL#26).

BTW, and here is some tests (don't want to create separate topic):

3
CC
00
1
01
*****
ans=-1

3
BAB
10
111
0
*****
ans=7

3
BA
00
01
1
*****
ans=3

3
CA
001
000
01
*****
ans=5

Input:
3
'╣' like '[╣-I]'
'╣' like '╣'
'╣' like '_'
Output:
YES
YES
YES

24
12 12 12 12 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Right Answer is 6

. - empty cage
1 - chess knight
2 - cells under attack

8 | . . . . . . . .
7 | . . . . . . . .
6 | . . 2 . 2 . . .
5 | . 2 . . . 2 . .
4 | . . . 1 . . . .
3 | . 2 . . . 2 . .
2 | . . 2 . 2 . . .
1 | . . . . . . . .
___________________
    a b c d e f g h
    1 2 3 4 5 6 7 8

is it possible to solve this problem with "smart" backtracking algo using a lot of restrictions ?

Please I got Crash(Stack Overflow), I implemented my first heavy-light descomposition and tried this problem, but I got this.

I got AC, I had to  convert the DFS to BFS, to evade the StackOverflow.

Edited by author 02.11.2011 04:45

Edited by author 02.11.2011 07:20

Edited by author 02.11.2011 08:20

Sir Isaac Newton is important to this problem.
In TWO different ways.

Why WA 28?I don't understand.
Here is my code:
program gjkh;
var a,b,ch,n,i,kol,sh,fir,sec,thi:longint;
m:array[1..100001]of longint;
flag:boolean;
begin
read(a);readln(b);
n:=0;
while not seekeof do begin
read(ch);
  if (n=0)or(m[n]<>ch)then begin
  n:=n+1;
  m[n]:=ch;
  end;
end;
flag:=true;
kol:=1;
for i:=1 to n do begin
if (i=1)then fir:=m[i]
  else begin
    if (flag=true)then begin
    sec:=m[i];
    flag:=false;
    end
    else begin
    thi:=m[i];
      if ((thi-sec)*(sec-fir)>0)then begin
      fir:=sec;sec:=thi;
      end
      else begin
      fir:=thi;
      kol:=kol+1;
      flag:=true;
      end;
    end;
  end;
end;
writeln(kol);
end.

Edited by author 16.06.2021 15:21

Edited by author 16.06.2021 15:22

see https://github.com/yosupo06/library-checker-problems/issues/610

my AC does not pass this short test

#include<algorithm>
#include<cstdio>
using namespace std;
#define MAX_N 250000

int n, i, t1, t2;
int *heap;

int main(){
    scanf("%d", &n);
    heap = new int[n/2+2];
    for(i = 0; i < n/2+1; ++i){
        scanf("%d", heap+i);
        push_heap(heap, heap+i+1);
    }

    for(i = 0; i < n/2-(n+1)%2; ++i){
        scanf("%d", heap+n/2+1);
        push_heap(heap, heap+n/2+2);
        pop_heap(heap, heap+n/2+2);
    }

    if(n%2){
        printf("%d.0\n", heap[0]);
    }else{
        t1 = heap[0];
        pop_heap(heap, heap+n/2+1);
        t2 = heap[0];
        if((t1+t2)%2){
            printf("%d.5\n", t1/2+t2/2);
        }else{
            printf("%d.0\n", t1/2+t2/2+t1%2);
        }
    }
    delete[] heap;

    return 0;
}

#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
    int i,j,k,l,ii;
    int m,n;
    int girls[101];
    int boys[101];
    int count = 0;
    int upsets = 0, girlUpsets = 0, boyUpsets = 0, currentUpsets = 0;
    int happyGirls = 0;
    int chooseGirl = 0, chooseBoy = 0;

    for (i = 0; i <= 100; i ++)
    {
        girls[i] = 0;
        boys[i] = 0;
    }

    cin >> n >> m; //n girls m boys
    for (i = 1; i <= n; i ++)
    {
        cin >> girls[i];
    }
    for (j = 1; j <= m; j ++)
    {
        cin >> boys[j];
    }

    count = min (m, n);
    //计算无配对upset值
    for (i = 1; i <= n; i ++)
        upsets += girls[i];


    happyGirls = 0;


    for (i = 1; i <= count; i ++)

    {
        for (j = 1; j <= n; j ++)
            for ( k = 1; k <= m; k ++)
            {
                girlUpsets = 0;
                boyUpsets = 0;
                //计算girl upset值
                for (l = 1; l <= n; l ++)
                {
                    if (l != j)
                    {
                        girlUpsets += girls[l];
                    }
                }


                //计算boy upset值
                for (l = 1; l <= m; l ++)
                {
                    if (l != k)
                        boyUpsets += boys[l] * (happyGirls + 1);
                    //cout << "werqe: " << boys[l] << '\t' << boyUpsets << endl;
                }


                currentUpsets = girlUpsets + boyUpsets;
                if (currentUpsets < upsets)
                {
                    upsets = currentUpsets;
                    chooseGirl = j;
                    chooseBoy = k;
                }

            }

        if (chooseBoy != 0  && chooseGirl != 0)
        {
            happyGirls ++;

            //删除第chooseGirl个girl
            for (ii = chooseGirl; ii <= n - 1; ii ++)
               girls[ii] = girls[ii + 1];
            n --;

            //删除第chooseBoy个boy
            for (ii = chooseBoy; ii <= m - 1; ii ++)
                boys[ii] = boys[ii + 1];
            m --;
        }

    }
    cout << upsets << endl;
    return 0;
}