If you think it in a way of dynamic programming it could help.
Imagine we have N pirats. Calculate probability of draw for single round (either all N pirats has the same gesture or there's a pirat for every possible gesture).

Then calculate average number of rounds to be played before someone lost (standard geometric sum).

Now someone is lost and we have the same game but with less pirats. So we can use DP-approach with probabilities.

Use doubles, it's enough to get AC.
You would be happy to implement this problem (my impl is 30 lines with i/o on c++).

I just used some identities and solved some equations to get the answer. Pretty sure I wouldn't be able to solve if I did not have access to the net and that is pretty lame :/

Was anybody able to solve by deriving the solution themselves? If yes, then please share! I want to learn the approach! ^_^

Nice problem.
Playing out with small numbers and sample could help in sequence instantiation.

My solution is only 30 lines of c++ code O(L) though.

Hint: imagine we have sequence which can replace up to L spaces afterall. Then to enlarge our sequence we insert L/2 + 1 in the beginning. Maximum L could be computed via straightforward iteration starting from lax max_L
Example:

seq: 2
max_L: 2

seq: 2, 2/2 + 1 = 2, 2
max_L: 2, 4

seq: 2, 2, 4/2 + 1 = 2, 2, 3
max_L: 2, 4, 10

seq: 2, 2, 3, 10/2 + 1 = 2, 2, 3, 6
max_L: 2, 4, 10, 40

Etc.
GL.

1
23:59:59

->

00:02:47

people! who has any ideas?

Can anyone explain why this is saying wrong answer?

namespace Ex2
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] input = Console.ReadLine().Split(new char[] { ' ', '\t' }, StringSplitOptions.RemoveEmptyEntries);

            for (int a = input.Length - 1; a >= 0; a--)
            {
                double temp = Math.Sqrt(double.Parse(input[a]));

                if (temp == 0)
                {
                    string zero = temp.ToString();
                    zero += ",";
                    Console.WriteLine(zero.PadRight(6,'0'));
                }
                else
                {
                    Console.WriteLine(temp);
                }
            }

            //Console.ReadLine();
        }
    }
}

import java.util.Scanner;

public class main {
    public static void main(String[] args) throws Exception {

        Scanner scan = new Scanner(System.in);

        String a = scan.nextLine();
        double b = Double.parseDouble(a);
        double c;
        double d = b;

        double mass[] = new double[(int)b];

        while (b != 0) {
            c = Math.sqrt(scan.nextDouble());
            mass[(int)(b - 1)] = c;
            b--;
        }
        for (double i = 0; i < d; i++) {
            System.out.printf("%.4f\n", mass[(int)i]);
        }
    }
}

Outputs "Runtime error"

Edited by author 09.10.2017 00:50

You can build binary tree
Left child corresponds to inward fold
Right child corresponds to outward fold
Root is unfolded strip
Build tree from root to leaves and before going deeper check winning conditions

How to deal with case like this
' ' like ' like ' like ' like ' 'like '
it can be splitted as
( ) & ( like ' like ' like ' 'like ) or
( ' like ) & ( like ' like ' 'like ) or
( ' like ' like ) & ( like ' 'like ) or
( ' like ' like ' like ) & ( 'like )

Can you clarify this statement
Inner entrance of apostrophe symbol (ASCII 39) into string or template is encoded by double apostrophe symbol

And if it's true the case
''hi'' like ''hi''
will be changed to
'"hi'' like '"hi'
Am I right?

Then how is this case possible?
'''''' like '_'''

Does anyone know about test number 8?. If you do, please upload some, thank you very much.

#include <stdio.h>
int main()
{
    int a,b;
    scanf("%d,%d",&a,&b);
    printf("a+b=%d\n",a+b);
    return 0;
}

please, how can I get test cases on this oj?

Error in jury solution was fixed, all test answers updated, few new tests added. Most authors lost their AC.
Big thanks to the Winger for pointing out an issue and helping with fixing it.

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int a, a2, a3;
    int b, b2, b3;
    scanf("%d %d %d %d %d %d", &a, &b, &a2, &b2, &a3, &b3);
    int bb1 = a - a3;
    int bb2 = b - b2;
    printf("%d %d\n", bb1, bb2);

    return 0;
}

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#define SIZE 128*1024
int main()
{
    int i,n=0;
    long long int num[SIZE],N;

printf("enter your aray value:\n");
    while(scanf("%lld",&N)!=EOF)
          {
              num[n]=N;
            n++;
        }

    for(i=n-1;i>=0;i--){

      printf("%0.4lf\n",(double)sqrt(num[i]));
    }
    return 0;
}

4th test is:

3
0 0
9999 10000
9998 9999

from re import *
from math import sqrt

result = findall('[0-9]+', input())
result.reverse()

for i in result:
    print('{:.4f}'.format(sqrt(int(i))))

Edited by author 14.03.2017 01:20

Who had a Wrong Answer in 5 th test ? What mistake may be there ?

Could someone provide tests for me? Thank you

Try this test:
4 3
####
#..#
####
Answer is 1 :)