I'm newby here and I didn't solve this kind of tasks before.
I know, that must be very easy, give me please some pieces of advice to solve it.

Hey, guys, it is not serious, really ...... I had wa4 with EPS = 1e-8 .... when I've changed EPS = 1e-6 I got AC...

Edited by author 06.02.2014 18:13

Does anyone know what's this test like?

using System;
namespace ConsoleApplication4
{
    class Program
    {
        static void SQRT(UInt64 sum)
        {
            double ima_ans = Convert.ToDouble(Math.Sqrt(Convert.ToDouble(sum)));
            Console.WriteLine(Convert.ToDouble((Math.Round(ima_ans = ima_ans * 10000)) / 10000));
        }
        static void Main(string[] args)
        {
            string save_str=null;
            string str = Console.ReadLine();
            int i = str.Length;
            int  k;
            UInt64 sum;
            i = i - 1;
            while (i >= 0)
            {
                k = 0;
                while (i >= 0 && str[i] != ' ')
                {
                    k = 1;
                    save_str += str[i];
                    i = i - 1;
                }
                sum = Convert.ToUInt64(save_str);
                if (sum > 0 && sum <= Math.Pow(10, 18) || k == 1)
                    SQRT(sum);
                save_str = null;
                i = i - 1;
            }
            Console.ReadLine();
        }
    }
}

Don't have any idea, give me that or similar test please.

#include <iostream>

using namespace std;
int main()
{
    int a,b;
    cin>>a;
    if(a==0 && a<=10000)
    cout<<"1";
    else
    {
    if(a>0 && a<=10000)
    cout<<((a*(a+1))/2);
    else
    { if(a<=10000){
        b=-((((-a)*((-a)+1))/2)-1);

        cout<<(b);}
        else
        cout<<"error";}

    }
return 0;
}

Edited by author 05.02.2014 01:45

Ok, just search biggest prime number :)

Edited by author 04.02.2014 22:13

more than one registrations ;)

who can help me? I WA on #7.Why?

here is my program:
var
  a:array[1..1000]of longint;
  b:array[1..1000,1..1000]of boolean;
  n,i,j,x,y,m:longint;
  c:boolean;
  function find(x,y:longint):boolean;
    var
      i:longint;
    begin
      if b[x,y] then exit(false);
      for i:=1 to m do
        if (b[a[i],x])and(b[a[i],y])then
          begin
            b[x,y]:=true;b[y,x]:=true;
            exit(false);
          end;
      exit(true);
    end;

begin
  while not seekeof do
    begin
      readln(x,y);
      b[x,x]:=true;b[y,y]:=true;
      b[x,y]:=true;b[y,x]:=true;

      c:=true;
      for j:=1 to m do
        if a[m]=x then begin c:=false;break;end;
      if c then
        begin
          inc(m);a[m]:=x;
          for i:=1 to m do
            if find(a[i],x) then
              begin
                writeln(0);halt;
              end;
        end;

      c:=true;
      for j:=1 to m do
        if a[m]=y then begin c:=false;break;end;
      if c then
        begin
          inc(m);a[m]:=y;
          for i:=1 to m do
            if find(a[i],y) then
              begin
                writeln(0);halt;
              end;
        end;
    end;
  writeln(1);
end.

[code deleted]

Edited by author 10.09.2012 19:45

Edited by moderator 20.11.2019 00:11

n = 3
  1  2  0  0
  2  3  0  0
  0  3  1  0

n = 6
  1  2  3  0  0  0
  3  4  5  0  0  0
  0  5  6  1  0  0
  0  0  6  2  4  0

n = 10
  1  2  3  4  0  0  0  0
  4  5  6  7  0  0  0  0
  0  7  8  9  1  0  0  0
  0  0  9 10  2  5  0  0
  0  0  0 10  3  6  8  0

n = 15
  1  2  3  4  5  0  0  0  0  0
  5  6  7  8  9  0  0  0  0  0
  0  9 10 11 12  1  0  0  0  0
  0  0 12 13 14  2  6  0  0  0
  0  0  0 14 15  3  7 10  0  0
  0  0  0  0 15  4  8 11 13  0

n = 21
  1  2  3  4  5  6  0  0  0  0  0  0
  6  7  8  9 10 11  0  0  0  0  0  0
  0 11 12 13 14 15  1  0  0  0  0  0
  0  0 15 16 17 18  2  7  0  0  0  0
  0  0  0 18 19 20  3  8 12  0  0  0
  0  0  0  0 20 21  4  9 13 16  0  0
  0  0  0  0  0 21  5 10 14 17 19  0

n = 36
  1  2  3  4  5  6  7  8  0  0  0  0  0  0  0  0
  8  9 10 11 12 13 14 15  0  0  0  0  0  0  0  0
  0 15 16 17 18 19 20 21  1  0  0  0  0  0  0  0
  0  0 21 22 23 24 25 26  2  9  0  0  0  0  0  0
  0  0  0 26 27 28 29 30  3 10 16  0  0  0  0  0
  0  0  0  0 30 31 32 33  4 11 17 22  0  0  0  0
  0  0  0  0  0 33 34 35  5 12 18 23 27  0  0  0
  0  0  0  0  0  0 35 36  6 13 19 24 28 31  0  0
  0  0  0  0  0  0  0 36  7 14 20 25 29 32 34  0

Thanks to author!

Edited by author 19.11.2014 21:49

Maybe, you know some tricks?

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

int main(void)
{
    long long number[1000000];
    int i=0,k;
    while(scanf("%lld", &number[i++]) == 1);

    k=i-1;

    while(k>0)
    {
         printf("%.4lf\n", sqrt((double)number[--k]));
    }

    return 0;

}

#include<stdio.h>
#include<math.h>
unsigned long long int a,i,s[10000000];
int main()
{
    double b;
    i=0;
    //freopen("input.txt","r",stdin);
    while(scanf("%llu",&a)!=EOF)
    {
        s[i]=a;
        i++;


    }
    while(i>0)
    {
        i--;
        b=sqrt(s[i]);
        printf("%.4lf\n",b);
    }

    return 0;
}

Can`t understand what is the problem

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

int main()
{
    unsigned long long int a[4096], b;
    int counter=0;
    while(scanf("%llu", &b))
    {
        a[counter]=b;
        counter++;
    }

    for(int i=counter-1; i>=0; i--)
        printf("%.15Lg\n", (long double)sqrt(a[i]));
    return 0;
}

Сan somebody tell whether it may be several current users at once?
Может кто нибудь сказать, может ли быть несколько текущих пользователей сразу?

Edited by author 01.02.2014 02:35

I tried to solve it with (dp[i] = sum_j(max(dp[j - 1], dp[i - j])) + 10) formula, but the way of getting the math expectation as max(dp[A], dp[B]) isn't right. Ho to do it? What is the formula..?

please, help me... :(

В FPC все работает, здесь же ...

Вот код, помогите
program f1;
var f,f2:text;
i,j:integer;
a:array[1..130000] of real;
begin
assign(f,'input.txt');
reset(f);
i:=0;
While not eof(f) do begin
                    inc(i);
                    read(f,a[i]);
                    end;
close(f);
assign(f2,'output.txt');
rewrite(f2);
For j:=i downto 1 do

writeln(f2,sqrt(a[j]):0:4);
close(f2);
End.

Do NOT use STL's strings, use standart C char arrays instead. That makes a HUGE difference in the time and memory complexity. I hope that this will be helpful to somebody:)