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:)

I tried to figure it out, how 1 tree and two sticks with the length of 2 can form the area more than 4? The largest area would be rectangle. What is the trick on this problem, can anybody explain with Geometry language.

Thanks.

Edited by author 30.01.2014 20:18

It keeps saying that my program produces incorrect solutions. Anyone else is having the same issue?

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

int main() {
  int start, change, N, sum;
  scanf("%d", &N);
  if (abs(N) <= 10000) {
    change = N > 1 ? 1 : -1;
    for (start = 1; start != (N + change); start += change)
      sum += start;
    printf("%d\n", sum);
  }
  return 0;
}