I decided to write post my solution and comment it, may be it will
be useful for someone.
So, you have recursively defined sequence. Of course first idea is
to write recursive function. But, we need to find max element from
1 to n and if you will call function for every n in order to find
maximum element, such algorithm will cost O(n^2). Another way is to
store results gotten before in special array and compare gotten result
with max of them.
I hope I helped you at least as it.
And here is accepted solution in C++:
#include <iostream>

#define max(a, b) ((a) < (b) ? (b): (a))

int last = 1;
int mem[100000];
int max[100000];

int F(int n)
{
    for (int i = last+1; i <= n; ++i) {
        if (i%2)
            mem[i] = mem[i/2]+mem[i/2+1];
        else
            mem[i] = mem[i/2];

        max[i] = max(mem[i], max[i-1]);
    }

    return max[n];
}

int main()
{
    mem[0] = 0; mem[1] = 1;
    max[0] = 0; max[1] = 1;

    int n;
    while (1) {
        std::cin >> n;

        if (!n) break;

        std::cout << F(n) << std::endl;
    }


    return 0;
}

I want to submit the solution in C language. What language I should choose from the option and what is the error in following coding:-
#include <stdio.h>

int main()
{
    double a,b;
    scanf("%lf %lf",&a,&b);
    printf("%lf",a+b);
    return 0;
}

Edited by author 31.03.2015 20:19

Try this test:
1 0 0
0
1
1 1 0

Answer: You should rent the apartment #1 alone.

Есть ли возможность сдавать задачи на языке Фортран

Each crossroads in the input is the binary OR of its branches, so a 5 (binary 101) at some crossroads in the input means there are branches 4 (100) and 1 (001) there.

As for the output, for each crossroads where there are no branches (each 0, let's call this crossroads P), repeat this:
Find the Manhattan (vertical+horizontal) distance to the closest branch from C. In the description, this closest distance is 2. Next, find all branches that lie on this exact distance from C. In the description, these are branches with number 16, 8, and 4. The output at location P should reflect all types of branches at distance 2 by ORing these numbers together, and in the description we get 28 (binary 10000 | 1000 | 100).

import java.util.Scanner;
import java.io.*;

public class HelloWorld {

    public static void main(String args[]) {

        int a = 0;
        int total = 0;
        Scanner input = new Scanner(System.in);
        a = input.nextInt();


            total = convertFunc(a);
            System.out.println(total);
    }

        public static int convertFunc(int n){

        int sum = 0;
        if(n > 0 && n <= 10000 )
        {
            for(int i = 1 ; i <= n ; i++ )
            {
                sum += i;
            }
        }
        else
        {
            for(int i = 1 ; i >= n ; i--)
            {
                sum += i;
            }
        }
        return sum;
    }
}

import java.text.DecimalFormat;
import java.text.DecimalFormatSymbols;
import java.util.ArrayList;
import java.util.Scanner;


public class Task1001
{
    public static void main(String[] args)
    {
        Scanner scanner = new Scanner(System.in);
        ArrayList<Long> list = new ArrayList<Long>();
        for (int i = 0; i < 4 ; i++)
        {
            long a = scanner.nextLong();
            if (a >= 0 && a <= 10E18)
            list.add(a);
        }
        DecimalFormatSymbols symbol = new DecimalFormatSymbols();
        symbol.setDecimalSeparator('.');
        DecimalFormat f = new DecimalFormat("#,##0.0000", symbol);

        for (int i = list.size(); i > 0 ; i--)
        {
            double c = (Math.sqrt(list.get(i - 1)));
            System.out.println((f.format(c)));
        }
    }
}



I need help please!!!

Edited by author 04.04.2015 17:02

Problem was solved.

Edited by author 04.06.2015 18:02

Please give me some tests

var a, b : array[1..1000] of integer;
    n, i, k : integer;
begin
  readln(n);

  for i := 1 to n do begin
    read(a[i]);
  end;
   i := 2;
   k := 1;
   while i <= n + 2 do begin
     if a[i] = a[i - 1] then begin
     k := k + 1;
     i := i + 1;
     end
     else
     begin
       write(k,' ',a[i - 1],' ');
       i := i + 1;
       k := 1;
     end;
   end;


end.

you need to write answer with two digits after point, so don't use cout!
use printf("%.2f\n",  );
and don't forget #include <cstdio>

i've been finding this mistake for 2 days!!

Does someone have any idea what's test 3?

i understand problem !!!

Edited by author 03.06.2015 12:14

если жарить 3 бифштекса на сковороде, где умещается всего 2, то потребуется не 3, а 4 минуты, поскольку последний бифштекс на 2 части не разделить и не положить на сковороду одновременно обоими сторонами

java

public class MainClass
{
    public static void main(String[] args)
    {
        System.out.println("A=\"'\";B='A=\"\";B=;?$(B,0,4)+A+$(B,4,4)+A+B+A+$(B,8,37)';?$(B,0,4)+A+$(B,4,4)+A+B+A+$(B,8,37)");
    }
}

This is my code
   #include <iostream>
   #include <stdio.h>
   #include <math.h>
   using namespace std;
long n,x[201],y[201];

   int main()
  {double xa,xb,xc,ya,yb,yc,l,a,b,c,x,h;
   cin>>xa>>ya>>xb>>yb>>xc>>yc>>l;
  b=sqrt((xc-xa)*(xc-xa)+(yc-ya)*(yc-ya));
  a=sqrt((xc-xb)*(xc-xb)+(yc-yb)*(yc-yb));
  c=sqrt((xa-xb)*(xa-xb)+(ya-yb)*(ya-yb));
  if(c!=0){
  x=(c*c+b*b-a*a)/(2.0*c);
  h=sqrt(b*b-x*x);
  if(b!=0&&a!=0){
  if((b*b+c*c-a*a)/(2.0*b*c)<=0) h=b;
  if((a*a+c*c-b*b)/(2.0*a*c)<=0) h=a;}

  if(h<=l) h=0; else h=h-l;}
  else h=a;
  printf("%0.2f\n",h);

  if(a>=b) {if(a<=l) h=0; else h=a-l;}
  else {if(b<=l) h=0; else h=b-l;}
  printf("%0.2f\n",h);


      return 0;
   }

Please give me some test

Edited by author 11.06.2015 18:48

Edited by author 11.06.2015 18:48

pls give me some test data, dont know what s wrong with code.
greetings.

My code:
a = readlines
a = a.join
a = a.split(/\s+/)
a.reverse!
a.map!{|elem|elem.to_i}
a.map!{|elem|(elem**0.5).round(4)}
puts a

I've tried code from FAQ page:
tokens = []
while string = gets do
   string.split().each do |token|
      tokens << token
   end
end
tokens.reverse.each do |token|
   puts (token.to_i ** 0.5).round(4)
end

Both variants return right results in my terminal, but in both cases i've got "wrong answer" here.
Why?