accept this kind of double:
#include <vector>
#include <iostream>
#include <algorithm>
#include <iterator>

#include <cmath>

using namespace std;

void root( double& d )
{
    d = sqrt(d);
}

int main()
{
    vector<double> arr;
    copy( istream_iterator<double>(cin), istream_iterator<double>(), back_inserter(arr) );

    for_each( arr.begin(), arr.end(), root );

    cout.precision(4);
    cout << fixed;
    copy( arr.rbegin(), arr.rend(), ostream_iterator<double>(cout, "\n"));

    return 0;
}

Wrong answer of "float" type:

#include <vector>
#include <iostream>
#include <algorithm>
#include <iterator>

#include <cmath>

using namespace std;

void root( float& d )
{
    d = sqrt(d);
}

int main()
{
    vector<float> arr;
    copy( istream_iterator<float>(cin), istream_iterator<float>(), back_inserter(arr) );

    for_each( arr.begin(), arr.end(), root );

    cout.precision(4);
    cout << fixed;
    copy( arr.rbegin(), arr.rend(), ostream_iterator<float>(cout, "\n"));

    return 0;
}

strange problem ......

 if (n==1||n==2)   what is  the  answer the subject?

This is my solution:

#include <stdio.h>

long double fact( long double n )
{
    if (n < 2)
        return 1;
    else
        return n * fact(n - 1);
}

int main( void )
{
    long double n, nfac, ans = 0, i;

    scanf("%Lf", &n);

    if (n < 2)
    {
        printf("0");
        return 0;
    }

    nfac = fact(n);

    for (i = 2; i < n; i ++)
        ans += nfac / fact(n - i);

    ans += nfac;

    printf("%.0Lf", ans);

    return 0;
}

WTF?

why

10
0.0
-0.0
1100000000000000000000000000000.0000000000
-0.0
24680976.32135864232426891300
Not a floating point number
Not a floating point number

is incorrect output for the sample (test#1)?

Edited by author 24.10.2008 23:22

What the test?
Sample working correct on my computer.

Edited by author 09.04.2009 18:13

    The test is 2 27
    I found that some AC solution output like that:
    a-z
    {-az
    ba-bz

    It means that the testdatas do not strong enough.
    Please add this.
    By the way,I guess the test 10 may be have some characters that do not in ['a'..'z','-']
    Please check it.
    At last,sorry for my poor English.
    Thank you.

New tests have been added. Accepted solutions have been rejudged. 295 authors have lost AC after rejudge.

Could someone give me some tests??

Can this problem be solved better than O(N^3)?

Give me test 2, please

Hello!
How must be organized input from console for a series of numbers, which quantity isn't known?
If we worked with file i/o, we could check for symbol, that means end of file.
And what is the condition to break input here?

It is allowed that the empty string matches several patterns. ???

I seriously think that sample output is screwed up. For example, the lowest line reads:

-1 1 4 - 1 4

But taking into account the proximity of the branch with ID=4
(bin: 100) and the branch with ID=1 (bin: 001) shouldn't all non-negative numbers in line equal 5( bin: 101), like

-1 5 5 -1 5 ?

Could someone make it clear for me? I would be very grateful;-)

I receive compilation error but in Borland C++ all ok.
What is problem?

#include<iostream.h>
#include<stdlib.h>
#include<string.h>

void main()
{
    unsigned long N, K, i, position;
    char * mas[30000];
    char * mas1[30000];
    char * mas2[30000];
    char * mas3[10000];

    cin>>N;
    cin>>K;


    for( i = 1 ; i <= N; i++)
    {
      if(i < 30000)
      {
        ultoa(i, mas[i - 1], 10/* size(i)*/);
      }
      else if (i < 60000)
      {
        ultoa(i,  mas1[i - 30001] , 10/* size(i)*/);
      }
      else if (i < 90000)
      {
        ultoa(i,  mas2[i - 60001] , 10/*size(i)*/);
      }
      else if (i < 100000)
      {
        ultoa(i,  mas3[i - 90001] , 10/*size(i)*/);
      }
    }

    i = 0;
    position = 0;
    int step = 0;
    while( i < N)
    {
        while(step < K)
        {
            if(position >= N)
                position = position - N;
            if(position < 30000)
            {
                if(strcmp(mas[position], "0") != 0)
                    step ++;
            }
            else if(position < 60000)
            {
                if(strcmp(mas[position - 30000], "0") != 0)
                    step ++;
            }
            else if(position < 90000)
            {
                if(strcmp(mas[position - 60000], "0") != 0)
                    step ++;
            }
            else if(position < 100000)
            {
                if(strcmp(mas[position - 90000], "0") != 0)
                    step ++;
            }
            position++;
        }
        step = 0;
        position --;
        if(position < 30000)
        {
            cout<<mas[position];
            mas[position] = "0";
        }
        else if(position < 60000)
        {
            cout<<mas1[position];
            mas1[position] = "0";
        }
        else if(position < 90000)
        {
            cout<<mas2[position];
            mas2[position] = "0";
        }
        else if(position < 100000)
        {
            cout<<mas3[position];
            mas3[position] = "0";
        }
        cout<<" ";
        i++;
        position ++;
    }
}

#include <iostream>
#include <stdio.h>
using namespace std;
int main()
{
    int i,n,a;
    double o=0;
    cin>>n;
    for(i=1;i<=n;i++)
    {
        cin>>a;
        o+=a;
    }
    o/=n;
    fprintf(stdout,"%.6lf ",o);
    cout<<endl;
    return 0;
}

Edited by author 07.04.2009 21:17

In English version of the site, the page offering to enter JUDGE_ID and password when viewing source has a button with Russian caption - "Продолжить"

Simple questiong :D
I just can't get it, how should we connect carriages?
If we have "AA" "BA" "AB"
what can we connect
"AABA" or "AAAB"???

#include <iostream>
using namespace std;
int main()
{
    int n,a[1024],br=1,i;
    cin>>n;
    cin>>a[1];
    if(n==1)
    {
        cout<<"1 "<<a[1];
    }
    for(i=2;i<=n;i++)
    {
        cin>>a[i];
        if(a[i]==a[i-1])
        {
            br++;
        }
        if(a[i]!=a[i-1])
        {
            cout<<br<<" "<<a[i-1]<<" ";
            br=1;
        }
        if(i==n)
        {
            cout<<br<<" "<<a[i];
        }
    }
    cout<<endl;
    return 0;
}

Edited by author 07.04.2009 21:12

program Ural1078;

var
  ok:array[1..500,1..500]of boolean;
  x,y,ans:array[1..500]of longint;
  a,b,n,max:longint;

function dfs(o,l:longint):boolean;

var
  i:longint;
  u:boolean;

begin
  u:=false;
  for i:=1 to n do
    if ok[o,i] then u:=u or dfs(i,l+1);
  if max<l then begin
    max:=l;
    u:=true;
  end;
  if u then ans[l]:=o;
  exit(u);
end;

begin
  readln(n);
  fillchar(ok,sizeof(ok),false);
  for a:=1 to n do
    readln(x[a],y[a]);
  for a:=1 to n do
    for b:=1 to n do
      if (x[a]<=x[b])and(y[b]<y[a]) then ok[a,b]:=true;
  max:=0;
  for a:=1 to n do
    if dfs(a,1) then ans[1]:=a;
  writeln(max);
  for a:=max downto 1 do
    write(ans[a],' ');
end.

What is wrong???

    I use greedy to solve this problem.
    First,I scanned all the boxes.If it had N new different coins that Tom wanted.The purchase will added about N*100
    Then,I think Tom would buy this box.And it costs him 100 yen.
    At last.If the purchase>0 ,then I would buy some boxes that Tom did not have been bought.
    I was WA on #1
    Am I wrong?
    At last,I'm sorry for my poor English.
    Thanks a lot.