I think this is a very hard case,and I don't know how to pass it....

- Do you know test 4 ?
- Any helpful tests ?
- If not, please pick bugs from my code. My algo: Find minimal, adjacent list, output sorted list. Tks a loooottttttt

Program Prufer_Code;
Type
        pnode = ^node;
        node = Record
         data:Longint;
         next:pnode;
End;

Var     adjList:Array[0..7500] of pnode;
        temp:pnode;
        A,C,minS:Array[0..7500] of Longint;
        n,i,j,s,p,q,m:Longint;
        fi:Text;
Procedure add(k:Longint;var x:pnode);
Var
        y,z:pnode;
Begin
        If (x=nil) then
         Begin
          new(x);
          x^.data:=k;
         End
        Else
         Begin
          new(y);
          y^.data:=k;
          y^.next:=x;
          x:=y;
         End;
End;

Begin

                n:=0;
                While not seekeof do
                 Begin
                  inc(n);
                  Read(A[n]);
                 End;

                For i:=1 to n+1 do
                 Begin
                  inc(C[i]);
                  inc(C[A[i]]);
                 End;
                dec(C[A[n+1]],2);
                p:=0;

                For i:=1 to n+1 do
                 If C[i]=1 then
                  Begin
                   inc(p);
                   minS[p]:=i;
                  End;

                For i:=1 to n do
                 Begin
                  q:=minS[1]; m:=1;
                  For j:=2 to p do
                   If minS[j]<q then
                    Begin
                     q:=minS[j];
                     m:=j;
                    End;
                  j:=q; minS[m]:=99999;


                  add(j,adjList[A[i]]);
                  add(A[i],adjList[j]);
                  dec(C[A[i]]);

                  If C[A[i]]=1 then
                   Begin
                    inc(p);
                    minS[p]:=A[i];
                   End;

                 End;

                For i:=1 to n+1 do
                 Begin
                  FillChar(C,sizeof(C),0);
                  temp:=adjList[i];
                  p:=0;
                  while temp<>nil do
                   Begin
                    inc(C[temp^.data]);
                    temp:=temp^.next;
                   End;
                  Write(i,': ');
                  For j:=1 to 7500 do
                   If C[j]=1 then Write(j,' ');
                  Writeln;
                 End;
End.

cin/cout with cin.sync_with_stdio(false) is 0.25 s, while scanf/printf 0.375 s.

Please, who solved this problem with disjoint sets send me your solution to the e-mail which is in my profile.
I wrote the code and it gives right answers on many tests, but not passes here. I need to compare where an error is.

It's very simple problem because all connected graphs which containes even number of edges can be divided into pairs of edges, which includes one common vertex!!!(and all connected graphs with odd number of edges can't be divided in such way)

I used this code:
double x, y; cin >> x >> y;
x *= 1000; y *= 1000;
int X = (int)x, Y = (int)y;

But this code doesn't work properly. For example, if x = -1.001, then X will be -1000 (in some cases one unit is lost).
How to avoid this in C++?
To solve this problem I had to read whole string and then parse it :)

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

int main(void){

    double a;
    int i = -1;
        double NUM [100000];

     while(EOF != scanf("%lf", &a)){
      NUM[++i] = sqrt(a);
     }

     i--;
     while(i>=0){
         printf("%.4f\n",NUM[i]);
         i--;
     }
     return 0;

}

long n=2*(value-1);
long sqr=(long) Math.floor(Math.sqrt(n));
if((sqr*(sqr+1))==n)
print(1);
else
print(0);

Edited by author 28.03.2014 15:49

Edited by author 28.03.2014 12:04

Is this a valid test case?

2
-9999 -1
-9999 1.00000000000000000001
1
9999 0

I know the answer should be 1. But it's hard to code.



Edited by author 08.03.2014 14:22

Edited by author 28.03.2014 23:31

aabaaab => { 0 1 0 1 2 3 }
a 0
aa 1
aab 0
aaba 1
aabaa 2
aabaab 3

When I used BFS,I found it would exceed the time limit.
Does DFS also will exceed the time limit?

Use probability P(i,j) that a line with i rockets is finished after at most j salvos. The probability that it ends after exactly k+1 salvos is

 P(i,k+1)-P(i,k) = 1/i sum_l P(l,k) * P(r,k) - P(l,k-1) * P(r,k-1),

where l is the index of the first launched rocket and thus the size of the left part and r is the size of the right part (r=i-1-l) of the line.

The output is with

 cout << setprecision(12) << s*10 << endl;

Edited by author 27.03.2014 23:04

I think my program is correct but i have not accepted yet.
why wrong answer #10..
please give me Test 10.

if a in [1..4] then write('few');
   if a in [5..9] then write('several');
   if a in [10..19] then write('pack');
   if a in [20..49] then write('lots');
   if a in [50..99] then write('horde');
   if a in [100..249] then write('throng');
   if a in [250..499] then write('swarm');
   if a in [500..999] then write('zounds');
   if a > 999 then write('legion');

Maybe that is what i am doing wrong.

If i have a set of words, say:

it
your
reality
real
our

Can the output be something like:

reality our reality

or the words can't be repeated??

Please give me some tests.

перепроверте тесты  задачи B pls

I think there can be several correct answers, and not all of them are accepted.

Is order of days important (Monday, Wednesday, Saturday == Saturday, Monday, Wednesday)?

For 3rd test my program prints "Monday, Tuesday, Wednesday" which gives 14 days. Why I got WA#3?

Thanks