All solutions now get WA1 / WA13, even those, which have been accepted earlier.
It's a judge error or just a rejudge coming?


Edited by author 22.09.2012 17:21

check this tests for "Not a floating point number"
-
1
-.
1
-.e0
1
-.1e
1
#123
1
--123
1
-1.e6
1

And last one (when I fixed this, I got Accepted):
-e20
1

Of course this is not an exhaustive list of possible bugs. Good luck, guys!

does anyone knows test 3? thank you!

import java.util.Scanner;
public class MEGA {

    public static void main(String[] args) {
        Scanner sc = new Scanner (System.in);
        int capacity = sc.nextInt();
        int minutes = sc.nextInt();
        int stoppedcars = 0;
        for (int i=1; i<=minutes; i++) {
            int thisminutecars = sc.nextInt();
            if (thisminutecars > capacity) stoppedcars = thisminutecars - capacity;
            if (thisminutecars < capacity) stoppedcars = stoppedcars - (capacity - thisminutecars);
        }
        if (stoppedcars < 0) stoppedcars = 0;
        System.out.println(stoppedcars);


    }

}

Edited by author 17.09.2012 17:51

I wrote an iterative primes generator using a 2-3-5 wheel for candidate generation and a priority queue for storing composite sequences.    I feel silly seeing all the ACs for trial division.   But at least now I know about wheel generators and iterative prime generation :).

program ural1078(input,output);
  type node=record x,y:longint;num:longint;end;
  var n,i,j,max,maxi:longint;
    f:array[1..500]of longint;
    c:array[1..500]of longint;
    l:array[1..500]of longint;
    a:array[1..500]of node;
  procedure sort(l,r:longint);
    var i,j:longint;k,temp:node;
    begin
      i:=l;j:=r;k:=a[(l+r)div 2];
      repeat
        while a[i].x>k.x do inc(i);
        while k.x>a[j].x do dec(j);
        if i<=j then begin
          temp:=a[i];
          a[i]:=a[j];
          a[j]:=temp;
          inc(i);dec(j);
        end;
      until i>j;
      if l<j then sort(l,j);
      if i<r then sort(i,r);
    end;
  procedure print(x:longint);
    begin
      if x=-1 then exit;
      print(c[x]);
      write(a[x].num,' ');
    end;
  begin
    readln(n);
    //if n=0 then writeln(0);
    for i:=1 to n do
      with a[i] do begin
      readln(x,y);num:=i;
    end;
    sort(1,n);
    fillchar(c,sizeof(c),255);
    for i:=1 to n do begin f[i]:=1;l[i]:=0;
      for j:=i-1 downto 1 do
        if (a[i].x<a[j].x)and(a[j].y<a[i].y) //Take Care!!!
           then begin
                  if f[j]+1>f[i]
                     then begin
                       f[i]:=f[j]+1;
                       c[i]:=j;
                     end;
                end;
    end;
    for i:=1 to n do
      if f[i]>max then begin
         maxi:=i;max:=f[i];
      end;
    writeln(max);
    print(maxi);
  end.

I've solved 1079,but this problem...Is any trick there?

Edited by author 01.09.2007 22:19

I submited several times and i got WA at test #8.
I really appreciate if you would be so kind by giving me test #8.
Thank you.

Edited by author 30.09.2005 16:11

Edited by author 30.09.2005 16:11

Hello,

I often want to brag about my accomplishments at acm.timus.ru, I want to show how many problems I have solved and that I'm one of the most geeky participant of Timus.

I very much like the way projecteuler encourages boasting, it has a profile with level and number of problems solved.

In the end, such a feature would benefit everyone: Timus users can boast, Timus itself gets more visibility and new participants.

# include <iostream>
# include <cstdio>

using namespace std;

const int Maxn = 130012;

int n, ans1, ans2, L1, L2, x, p[Maxn], q1, q2;

int GCD (int a, int b){

 if (b > a) return GCD(b, a);
 else
 if (b == 0) return a;
 else
return GCD(b, a % b);

}



int main () {
    freopen ("1.txt", "r", stdin);
    freopen ("2.txt", "w", stdout);

    cin>>n;

    for (int i = 1; i <= n; i++){
     cin>>x;
     p[x] = i;
    }

    q1 = 0;
    q2 = n + 1;
    ans1 = 0;
    ans2 = 0;

    for (int i = 1; i <= Maxn; i++){
        if (p[i] > 0) {
          q1 ++; q2 --;

          L1 = p[i] - q1; if (L1 < 0) L1 = -L1;
          L2 = p[i] - q2; if (L2 < 0) L2 = -L2;

          if (L1 > 0)
              if (ans1 > 0) ans1 = GCD(ans1, L1); else ans1 = L1;

          if (L2 > 0)
              if (ans2 > 0) ans2 = GCD(ans2, L2); else ans2 = L2;



        }

    }

    if (ans1 == 0) ans1 = n - 1; else ans1 --;
    if (ans2 == 0) ans2 = n - 1; else ans2 --;

    if (ans1 > ans2) cout<<ans1; else cout<<ans2;

return 0;
}

Edited by author 19.09.2012 17:23

1) число может быть с ведущими нулями
2) выводить надо с ведущими нулями(если они есть)

короче вот тест
1)
0

000000000002

2)
2
02

020000000089

AC code?
Put here..

my solution : for (i = 1; i<=n; i++) printf("%d ",i);
    for (i = n; i > 1; i--) printf("%d ",i);
had ac... but it's wrong!!

Edited by author 18.09.2012 18:57

Edited by author 18.09.2012 18:57

find out such x that 2^x < n < 2^(x+1)
2^x means 2*2*...*2 x times

x=log(n)/log(2);//c++


the maximum value of a[i] i=1,2,...,2^x is
max=f(x);
where f(x) is the fibonacci sequence
f(0)=1 f(1)=1 f(n)=f(n-2)+f(n-1)

so i just have to look if there is a larger number from 2^x to n

it is easy to find out the value of a[i] without knowing a[1]...a[i-1]

if l=2*k and r=2*p
m=(l+r)/2
a[m]=a[l]+a[r];

it is rather easy to prove that
a[2^x]=1;
a[2^(x+1)]=1;

so you just have to do a binary search for i from 2^x to 2^(x+1) calculating the values for the left and right border

this is the function that does exactly that //c++
int b(long l,long r,long i,int v1,int v2)
{
long m=(l+r)/2;
if (i==m)    return (v1+v2);
if (i<m)    return j(l,m,i,v1,v1+v2);
        return j(m,r,i,v1+v2,v2);
}


I DON'T UNDERSTAND WHY, BUT WHEN I USE MATH.H I GET COMPILATION ERROR


good luck !!!

Hi Experts,

           Can you explain the sample test case in problem?
Thanks in advance
Anupam

try this test
2
11.02.2000 12.02.2000 11.02.2001
09.02.2001 11.02.2001 10.02.2002
-----------
1