Dear Team,
It is really very good support by you.
But please, look at the problem 1001-in java.
There might be problem in compiler-there is no one who solved this question in java.
I will include my answer too.:

import java.util.Scanner;
public class second{
public static void main(String[] args) {
Scanner s = new Scanner(System.in);

double[] arr2=new double[200000];

int i=0,k=0;
do{
arr2[k]=s.nextDouble();
k++;
} while(s.hasNext());

for(i=k;i>0;i--){
System.out.printf("%.4f\n",Math.sqrt(arr2[i-1]));
}
}
}

My solution use emulation with some heuristic conditions (for avoiding deadlock). Solution is o(n).

Some test:
1
output
3
0 2 1
--------
2
output
8
1 3 4 2 0 1 3 2
--------
3
output
15
2 4 5 3 1 0 2 4 6 5 3 1 2 4 3
--------
4
output
24
3 5 6 4 2 1 3 5 7 8 6 4 2 0 1 3 5 7 6 4 2 3 5 4
--------
5
output
35
4 6 7 5 3 2 4 6 8 9 7 5 3 1 0 2 4 6 8 10 9 7 5 3 1 2 4 6 8 7 5 3 4 6 5

    Please tell me.
    Thanks.

#include <iostream>
using namespace std;
#include <math.h>

struct Point
{
    long double x;
    long double y;
};
struct Point2
{
    int number;
    long double angle;
};
void insertSort(Point2 ** a, long size) {
  Point2 * x;
  long i, j;

  for ( i=0; i < size; i++) {  // цикл проходов, i - номер прохода
    x = a[i];

    // поиск места элемента в готовой последовательности
    for ( j=i-1; j>=0 && (a[j]->angle < x->angle); j--)
      a[j+1] = a[j];      // сдвигаем элемент направо, пока не дошли

    // место найдено, вставить элемент
    a[j+1] = x;
  }
}
/*
int compare( const void *arg1, const void *arg2 )
{
   if(((Point2*)arg1)->angle>((Point2*)arg2)->angle)
       return -1;
   else
        return 1;

}*/
int main()
{

    int n;
    cin>>n;
    Point * arr = new Point[n];
    long double min_a=0;
    int min_ai;
    for(int i = 0;i<n;++i)
    {
        cin>>arr[i].x;
        cin>>arr[i].y;
        if((!i)||arr[i].y<min_a)
        {
            min_a = arr[i].y;
            min_ai = i;
        }
    }
    cout<<min_ai+1<<" ";
    Point2 ** arr2 = new Point2*[n-1];
    int j = 0;
    long double xx = arr[min_ai].x;
    long double yy = arr[min_ai].y;
    for(int i = 0;i<n;++i)
    {
        if(i!=min_ai)
        {
            arr2[j] = new Point2;
            arr2[j]->number = i;
            arr2[j]->angle =  (arr[i].x - xx)/sqrt(pow(arr[i].x-xx,2)+pow(arr[i].y-yy,2));
            ++j;

        }
    }
    //qsort(arr2,n-1,4,compare);
    insertSort(arr2,n-1);
    cout<<arr2[(n-1)/2]->number+1<<endl;
    return 0;
}

I don't know why, but qsort sorts them wrongly (I commented part that is qsort and changed it with simple O(n^2) sorting and got AC... But could tell me, what's so wrong with this qsort? I think I could get a better time with it.

The only email I found on this is site is timus_support@acm.timus.ru, which when I send an email to, I got mail return:

The original message was received at Sun, 24 Jan 2010 15:38:57 +0500
from mail-pw0-f57.google.com [209.85.160.57]

  ----- The following addresses had permanent fatal errors -----
judge_mailbox@busin.usu.ru
   (reason: 550 5.1.1 User unknown)
   (expanded from: <timus_support@acm.timus.ru>)

  ----- Transcript of session follows -----
550 5.1.1 judge_mailbox@busin.usu.ru... User unknown

Final-Recipient: RFC822; timus_support@acm.timus.ru
X-Actual-Recipient: RFC822; judge_mailbox@acm.busin.usu.ru
Action: failed
Status: 5.1.1
Diagnostic-Code: X-Unix; 550 5.1.1 User unknown
Last-Attempt-Date: Sun, 24 Jan 2010 15:39:05 +0500


---------- Forwarded Message ----------
From: Nat Pavasant <n...@gmail.com>
To: timus_support <timus_support@acm.timus.ru>
Date: Sun, 24 Jan 2010 17:38:44 +0700
Subject: ...

Very beautiful problem and very easy to solve it using simple geometric methods of working with vectors and points.

Pay attention to output:
In the first three lines output the coordinates of the points put INSIDE the triangle. In the following three lines output the coordinates of the points put OUTSIDE the triangle.

Edited by author 15.02.2010 18:39

#include <iostream>
#include <math.h>

using namespace std;

int main()
{
    int n,s;
    cin>>n;
    if(n%10==7)
        s=int((n-1)*log((double)2)/log((double)10));
    else
        s=int(n * log((double)2)/log((double)10));
    cout<<s<<endl;
    return 0;
}

What is lacking in terms? What could I forget?

My old accepted program does not pass some tests. Maybe it concerns only my solution, however it would be quite good to add these tests.

With recent changes to statement sample became wrong; I think sample input should now look like:

BID 0.01
BID 10000.00
BID 5000.00
BID 5000.00
SALE 7000.00 3
DEL 5000.00
SALE 3000.00 3
SALE 0.01 3
QUIT

, am I right?

I think, that tests for this problem are bad.
For test 3 3 3 2 1 my AC program gives answer 2, while answer 1 is correct.
P.S. My first AC solution is incorrect, but i hope, that my second AC solution is correct.

My program worked this tests, but WA2?
14
1 1 3 3 0 2 2 2 2 0
2 2 6 1 -1 1 5 6 5 0
2 2 0 3 -1 1 5 6 5 0
2 2 -4 0 -1 1 5 6 5 0
2 2 6 7 -1 1 5 6 5 0
2 2 5 7 -1 1 4 6 5 0
1 1 6 6 2 2 3 3 4 4
1 1 6 6 2 4 3 3 4 2
1 1 2 2 -1 -1 -1 -2 -2 -1
1 1 3 3 4 6 2 2 6 4
1 1 3 3 4 6 1 2 6 4
1 1 3 3 4 6 1 2 7 4
1 1 2 2 1 3 3 3 3 1
1 1 1 5 0 8 1 2 3 2

Answer:
4.576491
5.019765
5.398346
6.324555
10.676619
10.605551
7.071068
7.634414
1.414214
8.993230
8.993230
9.162278
1.414214
5.841619

Remember, that plain text can contain ANY symbols (digits, signums), except ( or )
(123)123(123) - YES

-5/7
+
-5/7
######
-1 3/7

0
/
-1 25/31
######
0

1/7
*
-49 7/9
#####
-7 1/9

-11 1/8
/
-11 1/8
#####
1

Test #14 is the first test where number 10000 appears

Edited by author 28.05.2011 00:10

If the only method possible is cheating, what is the point in giving such problems. It doesnt test anything.

In the first string of condition: "For example, 2^2−1 — the first Mersenne prime".
So, the first Mersenne prime is 2^2-1 = 3.

Then, in the sample we read:
18   -   3217

But, if so, the first Mersenne prime is 2, not 3, so, for the test

1
1

the program should output 2, not 3.

var
 sum,ans,a,b,c,a1,b1,c1,n:longint;
 i,j:longint;
begin
 readln(n);
 sum:=1;ans:=0;a:=2;b:=3;c:=4;a1:=1;b1:=1;c1:=1;
 for i:=1 to n do
  begin
   a1:=a1*a;
   a1:=a1 mod 1000;
   b1:=b1*b;
   b1:=b1 mod 1000;
   c1:=c1*c;
   c1:=c1 mod 1000;
  end;
 sum:=sum+a1+b1+c1;
 while sum<>0 do
  begin
   if sum mod 10=0 then inc(ans);
   sum:=sum div 10;
  end;
 writeln(ans);
end.

Rus: Покупатели заявляют цену, по которой они готовы купить (от 0.01 до 10000.00 бибриков).
Eng: Customers make their bids: announce a price at which they are ready to buy a pig (the price is between 0.01 and 10000.00 bibriks and always has exactly 2 digits after the decimal point).

There is no analog of "and always has exactly 2 digits after the decimal point" in Russian statement which significantly complicate solution.

Can you please tell me is this array correct
{0,5392,1890,84,6520,3149,2416,2835,80,8614,742,7696,6823,
9492,7710,9444,510,118,6522,3213,4499,6178,4565,763,1071,
8875,2688,9145,1211,9480,4056,1817,8661,5467,3358,2892,
2205,8691,1963,2386,8401,1047,3691,6824,825,7728,6797,1720
8194,9901,2823,1952,9344,5022,1421,6116,4511,1289,2133,
7494,7298,5012,9638,8753,5968,4029,4804,9556,924,1497,5886
6078,2085,3876,268,2910,8962,2970,1015,3931,1103,4872,4054
346,1119,931,4454,6530,1722,4266,9888,7961,2891,885,4461,
7731,3316,2155,93,2871,9710}
q[i]=f(i*10^6),i from 0 to 100

Edited by author 18.12.2006 12:52

Edited by author 18.12.2006 12:53