Are multiple '1' digits possible in the circle ? If yes , then will they produce separate click if they are currently in the window ?

Edited by author 05.11.2020 03:23

Edited by author 05.11.2020 03:23

Edited by author 05.11.2020 03:23

http://e-maxx.ru/algo/connected_components
You need to output number of components minus 1.
Tests:

4 4 0
3 4
1 2
3 1
2 4
0


1 0 0
0

6 0 0
5

6 5 1
1 2
1 4
2 5
3 6
4 5
1 2
1

6 5 0
1 2
1 4
2 5
3 6
4 5
1

6 3 4
1 2
2 3
4 5
2

6 4 0
1 3
3 4
4 6
5 6
1

Edited by author 04.11.2020 21:20

 longest palindromic substring is "kazak" itself. What am i missing ?

WHYYYYY
if u have same problem
solution is 3 4 5

The key to solve is that : Driver won't take less than offered, and Petr won't pay more than asked. So if you using a loop for example in each step a will be always min(a + b, c) and c will always be max(c - d, a)

What is test 23?

Very Easy Josephus's problem. Look it on geek for geeks

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