this my code:
import java.io.*;
import java.util.*;

public class Problem_1147 implements Runnable{
    public static void main(String []args){
        new Thread(new Problem_1147()).start();
    }
    public void run(){
        try{
            reader = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
            //reader = new StreamTokenizer(new BufferedReader(new FileReader("card.out")));
            //out = new PrintWriter(new FileWriter("card.out"));
            solve();
        }catch(IOException e){
            throw new IllegalStateException(e);
        }
    }

    final int COLOR_COUNT = 5501;
    final int N = 400001;
    int n;
    int width;
    int height;

    int xPos[] = new int[10001];
    int yPos[] = new int[10001];

    int color[] = new int[COLOR_COUNT];
    int x[]      = new int[1001];
    int y[]     = new int[1001];
    int z[]      = new int[1001];
    int t[]     = new int[1001];
    int cc[]    = new int[1001];
    int xc[] = new int[1009];
    int yc[] = new int[1009];

    short tColor[][] = new short[2002][2002];

    StreamTokenizer reader;

    PrintWriter out;

    int nextInt()throws IOException{
        reader.nextToken();
        return(int)reader.nval;

    }







    void solve()throws IOException{

        width = nextInt();
        height  = nextInt();
        n = nextInt();

        for (int i = 1; i<=n;i++){
            x[i] = nextInt();
            y[i] = nextInt();
            z[i] = nextInt();
            t[i] = nextInt();

            cc[i] = nextInt();
        }
        x[0] = 0;
        y[0] = 0;
        z[0] = width;
        t[0] = height;
        cc[0] = 1;

        Arrays.fill(xPos,0);
        Arrays.fill(yPos,0);

        for (int i = 0;i <=n ;i++)xPos[x[i]] = xPos[z[i]] = yPos[y[i]] = yPos[t[i]] = 1;



        for (int i = 1; i<= width; i++)xPos[i] += xPos[i-1];
        for (int i = 1; i<= height; i++)yPos[i] += yPos[i-1];

        for (int i = width; i >= 0  ; i-- ) xc[xPos[i]] = i ;
        for (int i = height; i>= 0  ; i-- ) yc[yPos[i]] = i ;

    /*    System.out.println("x coords ");
        for (int i = xPos[0] ; i<=xPos[width];i++)System.out.println(xc[i]);
        System.out.println("y coords ");
        for (int i = yPos[0] ; i<=yPos[height];i++)System.out.println(yc[i]);
    */
        for (int i = 0; i<=n;i++){
            int i_x = xPos[x[i]];
            int j_x = xPos[z[i]];
            int i_y = yPos[y[i]];
            int j_y = yPos[t[i]];
            short c = (short)cc[i];
        //    System.out.println(i_x +" " +j_x+" " +i_y+" " +j_y);
            for (int v = i_x; v < j_x; v++){
                for (int w = i_y; w  < j_y; w++){
                    tColor[v][w] = c;
                }
            }
        }

        Arrays.fill(color, 0);
        for (int i = xPos[0]; i < xPos[width]; i++){
            for (int j = yPos[0]; j < yPos[height]; j++){
                color[tColor[i][j]] += (xc[i+1] - xc[i])*(yc[j+1] - yc[j]);
            }
        }

        for (int i = 1; i<=2500;i++)if (color[i]>0)System.out.println(i+ " "+color[i]);
    }

}

var
  i,j,k,l,m:longint;
  n:qword;
  a:array[1..10000000]of qword;
  q:double;
begin
  n:=0;
  while not eof do begin
    inc(n);
    read(a[n]);
  end;
  for i:=n-1 downto 1do begin
    writeln(sqrt(a[i]):0:4);
  end;
end.

What answer for this test?
0 0 8 0 2 0 4 0 6 0

I've tried to solve this problem, using simple aproach, only one array in which i remember time until that part of the memory is allocated. On request I check if it's used, if is shift time to new value if not nothing change... On new allocation I search for the first free memory spot... Very simple, straight from the text, but I get WA on #4. Could someone post some tricky test or explain why my algorithm is wrong?

I don't know how to correct my program for more speedly work. Help anybody.
import java.util.*;
import java.math.*;

public class InvertedSqrt
{
    public static void main(String []args)
    {
    Scanner scan = new Scanner(System.in);
    ArrayList<BigInteger> input = new    ArrayList<BigInteger>();
    input.add(scan.nextBigInteger());
    while(scan.hasNextBigDecimal() && input.size() * Integer.SIZE <= 256 * 1024 * 1024)
    {
    input.add(scan.nextBigInteger());
    }
    scan.close();
    int i = input.size() - 1;
        while(i >=0)
        {
        System.out.printf("%.4f",java.lang.Math.sqrt(input.get(i).doubleValue()));
        System.out.print("\n");
        i--;
        }
    }
}

Hi.
I'm using :) such a construction in my code

struct maxY {
    bool operator () (const int& n1, const int& n2) {
              // compare something
    }
};

...

set<int, maxY> s;

in my VS 2009 it's ok, but Online Judge says it's a compilation error. Help to fix, please.

i need it

I have WA 3. what should i output for this test?
2 1
1.000000000001 0 0 0 0 0
0 0 0 -1 0 0

Please add the following test:
N=100, the following 10000 numbers -127.

Awhile there is no test where answer is negative.

In this problem,rounding the double value to the nearest integer appeares to be the most tricky part of the whole problem.
If you want to guarantee correct rounding, do it yourself with floor() and ceil().
At least,as it happened to me.:)

Edited by author 14.05.2010 21:41

Just change longint into int64.

It is not necessary a rectangle. Try

5
11111
11011
11011
10010
00000

Yes

and all its rotations.

BTW, to read you can use

#define F(i,n) for(int i = 0; i < n; ++i)
int n;
vector<string> s;
[...]
cin >> n;
s.resize(n, string(n,' '));
F(i,n) cin >> s[i];

and AC at 0.046 second.

I always crash on that test...float point wrong

Correct answer for this test:
----------
iT is HinT
----------
is
----------
It is hint
----------

#include<stdio.h>
#include<math.h>
int main()
{
     double a[100000];
     signed long int i = 0;
     while (scanf("%lf", &a[i]) != EOF)
     {
         i++;
     }
     i--;
     while ( i >= 0)
     {
         printf("%.4lf\n",sqrt(a[i]));
         i--;
     }
     return ;
}

What it could be? Any tests please.

Why I've got MLE(19 MB). I use KMP and I don't understand why at all! I appeal to you for help...
#include<iostream>
#include<string>
#include<vector>
using namespace std;
int main() {
  int  i, j, f=0, n;
  string what_to_find, gde;
  cin>>n>>what_to_find>>gde;
  what_to_find+="#";
  what_to_find+=gde;
  what_to_find+=gde;
  gde="";
  vector<short> next(what_to_find.size());
  next[0]=0;
  for(i=1; i<what_to_find.size(); i++){
    if(what_to_find[i]==what_to_find[next[i-1]])
      next[i]=next[i-1]+1;
    else {
      j = next[i-1];
      while (j>0 && what_to_find[i]!=what_to_find[j]){
      j=next[j-1];
      if(what_to_find[i]==what_to_find[j])
        next[i]=j+1;
      }
    }
    if(next[i]==n) {
      f=1;
      break;
    }
  }

  if(f) cout<<i-2*n;
  else cout<<-1;
  return 0;

}

подсмотренная ими у каких подвыпивших космонавтов
->
подсмотренная ими у каких-то подвыпивших космонавтов

If you use Java pay attention to size of arrays, that you used.
I used 2 arrays for saving current line and prev line of input(every has o(n) elements), and 2 arrays for saving result of calculating in prev Line, and for saving result of calculation in current line (every has o(n) elements).

Edited by author 21.02.2010 15:56