Let
0,1,2,..N-1     - outer door
N ..  2*N - 1   - 1-inner room doors
2*N .. 3N - 1   - 2-inner room doors
....
i*N  i*N + 1,  .. i*N + N-1   - i-th room doors

total  (K+1)*N  doors.

W[i][j] - minimum dist from i to j-th door, where  0 <= i,j < (K+1)*N

initially  W[i][i] = 0,  W[i][j] = inifinity  for i <> j.

read,  and set W[i][j] = 1   if there exists road from i to j doors.

-----------------
N times run following steps:

1. Floyd algorithm  for 0.. (K+1)*N - 1   doors.

2.
  for  1<= t <= K ,  0 <= i, j < N  update
        W[t*N + i][ t*N + j]  = min(W[t*N + i][t*N+j], W[i][j])
  Because  (t*N + i, t*N + j)  - is identical path as outer  (i,j) path.


Edited by author 31.10.2020 16:46

8 8
1 2 3 4 5 6 7 8
21

4 2
1 3
3

4 2
2 3
3

4 1
1
3

4 1
2
2

I have written the code the using user defined function thats why it has become much bigger.

#include<Stdio.h>
#include<math.h>
int Sn(int n);
int An(int a);
int main()
{
    int x;
    scanf("%d",&x);
    Sn(x);
}
int Sn(int n)
{
    int j;
    if(n==1) {An(1); printf("+1");}
    else { for(j=1;j<n;j++) printf("(");
        for(j=1;j<=n;j++)
       {if(j==1)
        {An(j);printf("+%d)",n);}
        else if(j!=n) {An(j);printf("+%d)",n-j+1);}
        else {An(j);printf("+1");}}}

}

 int An(int a)
{ int i,p;
    if(a==1)
   printf("sin(%d)",a);
   else
   {for(i=1;i<=a;i++)
       { p= pow(-1,i+1);
          if(i==1) printf("sin(%d",i);
          else if(p==1) printf("+sin(%d",i);
         else printf("-sin(%d",i);}
          for(i=0;i<a;i++) printf(")"); }
}

Edited by author 29.10.2020 21:19

Try this test:
input:
1
8
output:
1

Oddly enough, my solution printed "infinity" in this simple case, but was successful with first 35 tests

/*
* Author      : Arman Sykot
* Date & time : 28.10.2020 21:47:12 +06
*/

#include <bits/stdc++.h>
#define ll long long
#define the_flash ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)

using namespace std;

int main() {

    the_flash;

    int n;
    cin >> n;

    int arr[n][n];

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++)
            cin >> arr[i][j];
    }

    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= i; j++) {
            cout << arr[i - j][j] << " ";
        }
    }

    for (int i = n; i < 2 * n - 1; i++) {
        for (int j = n - 1; j >= i - (n - 1); j--)
            cout << arr[j][i - j] << " ";
    }

    cout << "\n";

    return 0;
}

If you are trying to flip signs on rectangles consider of using 1 request instead of many. :)

o_o

Edited by author 10.11.2020 12:55

Hi everyone!

If you are getting TLE in Test #2 or maybe another, I would like to give you a little help:

-> Test #2 seems to give large numbers (like 15000), and it gives you those numbers in increasing order. Think about it.

Now, if that isn't enogh, I can give you a little more help:

-> Create an array / vector / list which will store the prime numbers you find.

-> Create an algorithm that determines whether a number is prime or not (remember to only       compare with odd numbers smaller than the square root of the number from which you are trying to figure out its primality).

-> Create a function that calculates the n-th prime number. For that create a counter which increases everytime you find a new prime number and store that prime number in your array / vector / list.

-> Finally, remember the thing I said at the beginning of this post! :D

Hope it helps. :D

i don't understand why. maybe somebody could help...
here's the code

from math import sqrt
n=int(input())
l=0
a=[None]*n
b=[None]*n
for k in range(n):
    a[k]=int(input())
for z in range(n):
    p=0
    for i in range(n):
        if a[i]>p:
            p=a[i]
            l=i
    b[z]=n-1-l
    a[l]=-a[l]
for z in range(n):
    a[z]=-a[z]
while n>1:
     n=n-1
     a[b.index(n-1)]=2*sqrt(a[b.index(n)]*a[b.index(n-1)])
print('{:.2f}'.format(a[b[0]]))

Edited by author 27.10.2020 23:40

dp[N][N] - minimum number of added brackets, ok[l][r] - is [l, r] already solved, vector<pair<char, int>> add[N][N] - what and where brackets should be added to [l][r], to make it balanced.
make calc(l, r) function, if ok[l][r] then return dp[l][r],
if(l == r) then add[l][r] = {{reversed(s[l]), l}} dp[l][r] = 1
if(l + 1 == r and s[l] matches s[r]) then dp[l][r] = 0
if(s[l] == s[r]) you update dp[l][r] with dp[l + 1][r - 1]
or with dp[l][k] + dp[k + 1][r],
take care of already balanced strings

Precompute the results for ~200 numbers (at regular intervals) and hardcode them.
Then you can just bruteforces starting from the closest precomputed number.

Use 64 bit integer for resulting value. Initially I didn't believe result can be this big and spent a lot of time looking for another possible issues with code.

I am checking possibility of shot, I am checking edge case (when shot just touches the tank). My geometry calculations use epsilon (also tried to adjust epsilon, didn't help). Tried to output results with 7 decimals. I am using double. What else I could miss?

import java.util.Scanner;
import java.math.RoundingMode;
import java.text.DecimalFormat;
public class Misol1001{
public static void main(String[]args){
Scanner i=new Scanner(System.in);
DecimalFormat df = new DecimalFormat("#.####");
        df.setRoundingMode(RoundingMode.CEILING);
int a=i.nextInt();
long[] k=new long[a];
for(int j=0; j<a; j++){
k[j]=i.nextLong();
}

double[] l=new double[a];
for(int c=0;c<a; c++){
l[c]=Math.sqrt(k[c]);
}
for(int j=a-1; j>=0; j--){

System.out.println(df.format(l[j]));
}
}
}

I got wrong answer beacuse of test 6
Someone please explain what is test 6
Program in C




#include <stdio.h>
#include <math.h>

int main()
{
    long int n,m,y;
    int x,cont=0,aux1;

scanf("%d %d %d",&n,&m,&y);



for (int i = 0; i <=m-1 ; ++i)
{

    aux1=pow(i,n);

    if (aux1%m == y)
    {
        printf("%d ",i);
cont++;

    }


}
if (cont == 0)
{
    printf("-1\n");
}



    return 0;
}

Edited by author 13.10.2020 05:27

Use Bigmod Algorithm

В описании обоих примеров мы делаем одно и то же:
1. Расшифровали
2. Перевернули
3. Зашифровали
4. Показали.
Зачем нам данные о типе компьютера GOOD или BAD?

What is wrong with my code? :(

#include <stdio.h>
#include <stdlib.h>

int main (){

 int N;
 scanf ("%d",&N);

    for (int i = 0 ; i < N ; i++){
       char x;
       int y;
       getchar();
       scanf ("%c %d",&x,&y);
       int movimientos = 0;

       if (y + 2 <= 8)
        if (x - 1 >= 97){
         movimientos++;
        }

       if (y+2 <=8)
        if (x + 1<=104){
         movimientos++;
        }

       if (y-2 >= 1)
        if (x-1 >97){
         movimientos++;
        }

       if (y-2 >= 1)
        if (x + 1 <= 104){
         movimientos++;
        }

       if (x+2 <= 104)
        if (y - 1 >= 1){
         movimientos++;
        }

       if (x+2 <= 104)
        if (y+1 <= 8){
         movimientos++;
        }

       if (x-2 >= 97)
        if (y+1 <= 8){
         movimientos++;
        }

       if (x-2 >= 97)
        if (y-1 >= 1){
         movimientos++;
        }
       printf ("\n %d", movimientos);
       printf ("\n");
    }

 return 0;
}

Edited by author 21.10.2020 01:36

Edited by author 21.10.2020 01:37

Can anybody give me ideas about 21 test, please?

Edited by author 18.10.2020 22:11

Edited by author 19.10.2020 03:42