If I want to type name "de", and both names "dd" and "de" are exist in contacts, how much seconds it will take? Should i wait 1 second after first pressing 'd'-key?

I got problem in test#1. please help

Whos use => ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

=> Need to delete, for the solve this problem.
I think its a problem #CheckerMechanism

Even if in the statement it is said that: "Albanian dictionary, found exactly N different words S[i] in it", in test 24 there are identical words, so be careful at this particular case.

#include<iostream>
#include<math.h>
using namespace std;
int main()
{
    long long n,p,a=1;
    cin>>n;
    while(1)
    {
        p=(1-2*a+sqrt((2*a-1)*(2*a-1)+8*n))/2;
        if(2*n==2*p*a+p*(p-1))
        {
            cout<<a<<" "<<p;
            break;
        }
        a++;
    }
    return 0;
}

The first mistake was mentioned by people prior to me, and it can be easily corrected by understanding the sample I/O.
Anyway, the statement mentioned " Moreover, in the final position the upper side must be the same as it was in the initial position.", without emphasizing what shape the upper side may have.
So in this case, the sample I/O makes sense if and only if the upper side of the cube is a symmetrical shape.
Otherwise, you can make a cube by yourself, and draw an asymmetrical shape(e.g. enter shape) to simulate the sample, and you will find that it is IMPOSSIBLE to move 11 steps to satisfy the description.

const Nmax=500;
var
 Stx,Sty,xx1,yy1,xx2,yy2:array[1..Nmax] of real;
 numb1,numb2:array[1..Nmax] of integer;
 a:array[1..Nmax,1..Nmax] of boolean;
 Nnew:array[1..Nmax] of boolean;
 N,M,i,j,r,n1,n2:integer;
 ans:string;

procedure Peres(x1,y1,x2,y2,x3,y3,x4,y4:real);
var
 x0,y0:real;
 A1,B1,C1,A2,B2,C2,Det:real;
begin
 ans:='No';

 if(x1>x2) then begin
  x0:=x1;
  y0:=y1;
  x1:=x2;
  y1:=y2;
  x2:=x0;
  y2:=y0;
 end;
 if(x3>x4) then begin
  x0:=x3;
  y0:=y3;
  x3:=x4;
  y3:=y4;
  x4:=x0;
  y4:=y0;
 end;

 A1:=y1-y2;
 B1:=x2-x1;
 C1:=x1*y2-y1*x2;

 A2:=y3-y4;
 B2:=x4-x3;
 C2:=x3*y4-y3*x4;

 Det:=A1*B2-A2*B1;

 if(x3=x4)and(y3=y4) then begin
  if(A1*x3+B1*y3+C1=0)and((x1-x3)*(x2-x3)+(y1-y3)*(y2-y3)<=0) then ans:='Yes';
 end
 else begin
  if(Det<>0) then begin
    x0:=(C2*B1-C1*B2)/Det;
    y0:=(C1*A2-C2*A1)/Det;
    if((x1-x0)*(x2-x0)+(y1-y0)*(y2-y0)<=0)and((x3-x0)*(x4-x0)+(y3-y0)*(y4-y0)<=0) then ans:='Yes';
  end
  else begin
   if(A2*x1+B2*y1+C2=0) then begin
    if(x1<=x3)and(x3<=x2) then ans:='Yes';
    if(x1<=x4)and(x4<=x2) then ans:='Yes';
    if(x3<=x1)and(x1<=x4) then ans:='Yes';
    if(x3<=x2)and(x2<=x4) then ans:='Yes';
   end;
  end;
 end;
end;

procedure G(v:integer);
 var u:integer;
begin
 Nnew[v]:=false;
 for u:=1 to N do begin
  if(Nnew[u])and(a[v,u]) then begin G(u);end;
 end;
end;

BEGIN
 readln(N,M);
 for i:=1 to N do begin
 for j:=1 to N do begin
  a[i,j]:=false;
 end;
 end;

 for i:=1 to N do begin
  readln(Stx[i],Sty[i]);
  Nnew[i]:=true;
 end;

 r:=0;

 for i:=1 to M do begin
  readln(n1,n2);
  a[n1,n2]:=true;
  a[n2,n1]:=true;
  inc(r);
  xx1[r]:=Stx[n1];
  yy1[r]:=Sty[n1];
  xx2[r]:=Stx[n2];
  yy2[r]:=Sty[n2];
  numb1[r]:=n1;
  numb2[r]:=n2;

  for j:=1 to r-1 do begin
   Peres(xx1[j],yy1[j],xx2[j],yy2[j],Stx[n1],Sty[n1],Stx[n2],Sty[n2]);
   if(ans='Yes') then begin
    a[n1,numb1[j]]:=true;
    a[n1,numb2[j]]:=true;
    a[numb1[j],n1]:=true;
    a[numb1[j],n2]:=true;
    a[n2,numb1[j]]:=true;
    a[n2,numb2[j]]:=true;
    a[numb2[j],n1]:=true;
    a[numb2[j],n2]:=true;
   end;
  end;
 end;


 for i:=1 to N do begin
 for j:=1 to r do begin
  Peres(xx1[j],yy1[j],xx2[j],yy2[j],Stx[i],Sty[i],Stx[i],Sty[i]);
   if(ans='Yes') then begin
    a[i,numb1[j]]:=true;
    a[i,numb2[j]]:=true;
    a[numb1[j],i]:=true;
    a[numb2[j],i]:=true;
   end;
  end;
 end;

 G(1);
 ans:='YES';
 for i:=1 to N do if(Nnew[i]) then ans:='NO';
 write(ans);
END.

Are you sure you want tests with a lot of input data, such that scanf("%d",...) doesn't work on time?
I don't think this is okay. There is a read problem with new g++ compilators.

If anyone knows what test 2 is?

Assume each general's gold a[1],a[2],...,a[m]
divide them into b[1]&c[1],b[2]&c[2],...,b[m]&c[m],and b[1]+c[1]=a[1],b[2]+c[2]=a[2],...
(b[i] can be any partition from a[i])
Then sort b from big to small, sort c from small to big.
We can prove that between b[1]+c[1],b[2]+c[2],...,b[m]+c[m], the biggest minus the smallest would not be worse than the previous one. Continuously do this and we'll get the answer.
This solution is developed by Wenbin Tang. Thanks to him!

My solution in Go (library sort + O(N)) got TLE on 16. Wrote in in C++ and got accepted.
But I really surprised that someone (gadget, zalick, rozdestvenskiy) solved it in Python, which is slower then Go

Any one who can tell me why wa#5?

from sys import stdin
data = [int(x) for x in  stdin.read().split()]
dig = max(data[1:])
list ='1'+''.join(['1'+'0'*x for x in xrange(dig+1) if dig>sum(xrange (x))])
print ' '.join([list[x-1] for x in data[1:]])

I need you help. What could be the #5 test like?

My program gets WA #5. Did anybody have the same problem? Program works correctly with all inputs I have found.

Edited by author 09.09.2012 13:01

заходит решение
берем пары соседних чисел и выводим минимум из их нодов(

#include <iostream>
#include <vector>

using namespace std;

int gcd(int a, int b)
{
    if(a == b)
        return a;
    while( a != 0 && b !=0)
        if(a > b)
            a%=b;
        else b%=a;
    return a+b;
}
int main()
{
    int N;
    cin >> N;
    int array[1000];
    for(int i = 0; i < N; i++)
        cin >> array[i];
    if(N == 1) cout << array[0];
    else{
    vector <int> mins;
    for(int i = 0; i < N-1; i++)
        mins.push_back(gcd(array[i], array[i+1]));
    int min = mins[0];
    for(int i = 0 ;i < mins.size(); i++)
        if(min > mins[i])
            min = mins[i];
    cout << min;
    }
    return 0;
}
это вроде не камильфо!

Do not forget that you should compare solved tasks as ints, not as strings :)

Use list to make the adjacency list and not vector.

dp[i][j]=dp[j][k]+1 where a[j]*2=a[i]+a[k] can be done using a loop.
via: http://www.geeksforgeeks.org/length-of-the-longest-arithmatic-progression-in-a-sorted-array
time O(n^2) space O(n^2)