В FPC все работает, здесь же ...
Вот код
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.

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

using namespace std;

bool isOne(int n);
int main(int argc, const char * argv[])
{

    int n;
    cin >> n;

    int *nums = new int[n];
    for (int i = 0; i < n; i++) {
        int k;
        cin >> k;
        nums[i] = k;
    }

    for (int i = 0; i < n; i++) {
        int index = nums[i] - 1;
        if (isOne(index)) cout << 1 << " ";
        else cout << 0 << " ";
    }
    delete [] nums;
}

bool isOne(int n) {

    double ans = (sqrt((double) 8 * n + 1) - 1) / 2;
    int r = (int) ans;
    if (r * (r + 1) / 2 == n) return true;
    return false;
}

#include <iostream>
using namespace std;

bool isOne(int n);
int main(int argc, const char * argv[])
{

    int n;
    cin >> n;

    int *nums = new int[n];
    for (int i = 0; i < n; i++) {
        int k;
        cin >> k;
        nums[i] = k;
    }

    for (int i = 0; i < n; i++) {
        int index = nums[i] - 1;
        if (isOne(index)) cout << 1 << " ";
        else cout << 0 << " ";
    }
    delete [] nums;
}

bool isOne(int n) {
    int indexOf1 = 0;
    for (int i = 0; indexOf1 < n; i++) {
        indexOf1 += i;
        if (indexOf1 == n)
            return true;
    }
    return false;

}

Edited by author 26.01.2014 16:27

#include <iostream>
using namespace std;
int main(int argc, const char * argv[])
{
    int n;
    cin >> n;


    if (n >= 1) {
        cout << n * (n + 1) / 2;
    }
    else {
        cout << - (-n * (-n + 1) / 2) + 1;
    }

}

#include <iostream>
using namespace std;

int main(int argc, const char * argv[])
{

    int m, n;
    cin >> m >> n;

    int num = m * n - 1;

    if (num % 2== 1) {
        cout << "[:=[first]";
    }

    else {
        cout << "[second]=:]";
    }

}


num is total number of splits

Код:

#include <iostream>
#include <cmath>
#include <vector>

using namespace std;

int main( int argv, char *argc[] )
{
    long long n;
    vector<long long> v;
    while ( cin >> n )
    {
        v.push_back(n);
    }
    for ( int i = v.size() - 1; i >= 0; i--)
    {
        cout << sqrt(v[i]) << endl;
    }
    return 0;
}

wrong answer на первом тесте

Edited by author 29.12.2013 21:39

Hi
I had solved this problem with just 3 shortint...
It so easy look other messages in this problem to understand what is
happening!
All you need is:

    dec(n,n mod y);

where y is that K[i]`s which described in problem.
nothing is special for this problem...
With best Wishes
Aidin_n7@hotmail.com

"gunners" must be "guns" in "one integer n that is the number of gunners on board"

#include <iostream>
#include <fstream>
#include <cmath>
#define PI 3.14159265358979323846
using namespace std;

#ifndef ONLINE_JUDGE
    #define cin fin
    ifstream fin("input");
#endif

double N, R;

int main() {
    cin >> N >> R;
    N /= 2.0;

    if (R <= N) {
        cout << PI * R * R << '\n';
        return 0;
    }
    if (R >= N * sqrt(2.0)) {
        cout << 4.0 * N * N << '\n';
        return 0;
    }

    double A, B, C;
    A = N;
    C = R;
    B = sqrt(C * C - A * A);
    double angle = 90.0 - (acos(A / C) * 180.0 / PI) * 2.0;

    printf("%.3lf\n", 4.0 * (A * B + PI * R * R * angle / 360.0));
}

This program works fine on all the tests I saw on the forum. But I get WA #1 probably because of the wrong precision. I printed the value with 4 and 6 decimals but I get the same WA #1.
Could anyone tell me please how can I solve the "error" ? It's annoying to have a good solution and take WA.
Thanks a lot!


LATER EDIT:
It seems I succeeded with such a function:

inline void precise(double X, const int &decimals) {
    cout << (int)(X);
    cout << '.';
    X = X - (int)(X);
    for (int i = 1; i <= decimals; ++i) {
        X = X * 10.0;
        cout << (int)(X);
        X = X - (int)(X);
    }
    cout << '\n';
}

...

precise((double)(4.0 * (A * B + PI * R * R * angle / 360.0)), 3);

Have fun!

Edited by author 25.01.2014 04:19

#include<iostream>
using namespace std;
int main ()
{
    int n,a[500000],d,max=0,maxi;
    cin>>n;
    for (int i=0;i<n;i++)
    {
        cin>>d;
        a[d]++;
        if (a[d]>max)
        {
            max=a[d];
            maxi=d;
        }
    }
    cout<<maxi;
    return 0;
}

#include <iostream>
#include <stdio.h>
#include <vector>

using namespace std;

class Place {
public:
    char x;
    int y;

    bool isOnboard() {
        if (x < 'a' || x > 'h' || y < 1 || y > 8) return false;
        return true;
    }



    Place(char x, int y) {
        this->x = x;
        this->y = y;
    }

};



int main(int argc, const char * argv[])
{
    int N;
    cin >> N;


    vector<Place> places;

    for (int i = 0; i < N; i++) {
        char x;
        int y;
        cin.get(x);

        if (x == '\n') {
            i--;
            continue;
        }

        cin >> y;

        Place place (x, y);
        places.push_back(place);

    }


    for (int i = 0; i < N; i++) {
        Place place = places[i];
        int moves = 0;


        //left 2 up 1
        place.x -= 2;
        place.y -= 1;
        moves += place.isOnboard();

        //left 2 down 1
        place.y += 2;
        moves += place.isOnboard();

        //right 2 down 1
        place.x += 4;
        moves += place.isOnboard();
        //right 2 up 1
        place.y -= 2;
        moves += place.isOnboard();

        //right 1 up 2
        place.x -= 1;
        place.y -= 1;
        moves += place.isOnboard();

        //left 1 up 2
        place.x -= 2;
        moves += place.isOnboard();

        //left 1 down 2
        place.y += 4;
        moves += place.isOnboard();

        //right 1 down 2
        place.x += 2;
        moves += place.isOnboard();

        cout << moves << endl;

    }



}

Edited by author 24.01.2014 20:07

Hi,
I'm interested if there is solution for larger N like 1000 and even more?

Thanks.

Edited by author 25.03.2013 01:28

I'm using Floyd-Warshall algorithm, and my program correctcly passes different tests, but i have WA1. Why?

My program works true on test from conditions and on two my tests. Tell me please, what's a problem?

1)
3
2
1
2

2)
614
13
627
861
527
911
751
658
568
846
998
188
923
426
685

3)
1560
19
580
594
644
62
963
718
291
532
293
77
356
984
384
182
986
915
18
671
920

4)
555
27
816
170
523
595
471
313
970
877
465
50
819
517
721
143
394
872
905
462
346
639
193
902
945
927
133
401
410

5)
800
  15
  10
  20
  30
  40
  50
  60
  70
  80
  90
  100
  110
  120
  130
  200
  350

6)
1458
17
802
204
451
575
158
168
247
15
50
100
1000
257
354
125
358
125
268

Thanks.

Edited by author 22.06.2006 19:08

Edited by author 23.06.2006 12:20

Please, write there translated (RUSSIAN) version of this problem!!!! or you can send there for my adress   daminus_corp@mail.ru    > Thanks!!!!

Oh no WA#25! Please help me!

var
  a1,a2,a3,h,s,aa,bb:real;
  a:array[1..3,1..2] of integer;
  i,l:integer;


procedure readdata;
begin
  readln(a[1,1],a[1,2],a[2,1],a[2,2]);
  readln(a[3,1],a[3,2],l);
end;

procedure writedata;
begin
  if aa>=0 then
  writeln(aa:0:2) else writeln('0.00');
  if bb>=0 then
  writeln(bb:0:2) else writeln('0.00');
end;

function d(x1,y1,x2,y2:integer):real;
begin
  d:=sqrt(sqr(x1-x2)+sqr(y1-y2));
end;

procedure square;
var
  p:real;
begin
  a1:=d(a[1,1],a[1,2],a[2,1],a[2,2]);
  a2:=d(a[3,1],a[3,2],a[2,1],a[2,2]);
  a3:=d(a[1,1],a[1,2],a[3,1],a[3,2]);
  p:=(a1+a2+a3)/2;
  s:=sqrt(p*(p-a1)*(p-a2)*(p-a2));
end;

begin
  readdata;
  square;
  if (a[1,1]=a[2,1]) and (a[1,2]=a[2,2]) then
  begin
    aa:=d(a[1,1],a[1,2],a[3,1],a[3,2]);
    bb:=aa;
  end  else
  begin
  if (a1*a1+a2*a2<a3*a3) or (a2*a2+a3*a3<a1*a1) or (a1*a1+a3*a3<a2*a2) then
  begin
    aa:=d(a[1,1],a[1,2],a[3,1],a[3,2]);
    if d(a[2,1],a[2,2],a[3,1],a[3,2])<aa
    then aa:=d(a[2,1],a[2,2],a[3,1],a[3,2]);
  end  else
  begin
    h:=d(a[1,1],a[1,2],a[2,1],a[2,2]);
    aa:=s/(0.5*h);
  end;

  bb:=d(a[3,1],a[3,2],a[2,1],a[2,2]);
  if d(a[3,1],a[3,2],a[1,1],a[1,2])>bb then
  bb:=d(a[3,1],a[3,2],a[1,1],a[1,2]);
  end;
  aa:=aa-l;
  bb:=bb-l;
  writedata;
end.

Can anybody suggest me the 6th test case?