Looks like std::vector::rend() does not return const_reverse_iterator.
I have got CE on the line
for( std::vector< Spot >::const_reverse_iterator it = spots.rbegin(); it != spots.rend(); ++it )
but
for( std::vector< Spot >::/*const_*/reverse_iterator it = spots.rbegin(); it != spots.rend(); ++it )
was ok.

Здраствуйте!
Задача:
Написать такой класс А, чтобы данный фрагмент кода компилировался и работал.

А a1;
A a2=a1+2+A(3);

Код:

# include <iostream>

using namespace std;

class A{
      private:
      int x;
      public:
             A():x(0){};//Constructor
             A(int y):x(y){};
             A(const A & y)  //Copy constructor
             {
               x=y.x;
               return *this;
              }
             A operator+(A y) const;

      };

  A A::operator+(A y) const{
                   return A(x+y.x);
                           }

  int main()

  {
      A a1;
      A a2=a1+2+A(3);
      return 0;
  }

Test:
8
Ans
9 1

Hi.
#include<iostream>
#include<math.h>
#include<iomanip>
using namespace std;
int main(){
    int i=0;
    double *a=new double[8000000];
    double c;
    while(cin.eof()==false){
        cin>>c;
        a[i]=sqrt(c);
        i++;
    }
    i--;
    i--;
    while(i>=0){
        cout<<setprecision(4000)<<a[i]<<'\n';
        i--;
    }
    return 0;
}
WA 2. Why?

can anyone please tell me what to output if N=1 is given?

input1 : 05 26 2C 52 69 14 3F 31 4C 2A 69 65 1A 26 4B
input2 : 61 07 28 41 3B 63 07 2C 52 22 21 69 72 0B 42 5E
answer : 41 43 4D 20 49 43 50 43 20 4E 45 45 52 43 27 32
(P.S : I put the spaces only for that is easy to read && understand)
Why is first 2 bytes in answer is 41...?
Cuz, 61h xor 20h = 41h (20h -> 32d = ' ' = Space);
How to find the second,third,...?
41h xor 05h = 44h
44h xor 07h = 43h(Second 2 bytes);
43h xor 26h = 65h
65h xor 28 = 4Dh (Third 2 bytes);
Last some tips:
if you use algos for convert from Dec to Hex be careful;
your algo should convert
1 : 0d to 00h (not to 0h);
2 : 5d to 05h (not to 5h);
GoodLuck!!! ;-)

Edited by author 17.05.2009 08:41

Hi,

To those who are reading the input character by character, test 4 contains some Non Digit characters as well like new line or space at the end of input etc. In my program, I just put a condition to ignore those and then it worked fine.

Varun

I found this test on acmp.ru
21 1 16 11 5 18 21
My AC answer was 3 but real is 1
(21+21+5=47 && 18+16+11+1=46 => 47-46=1)

3
3 4 1 2
3 3 1 2
3 4 1 3

i run it and i see there's no differernce between mine and the accepted one in result.

#include <cstdlib>
#include <iostream>
#include <cmath>
#include <iomanip>

using namespace std;

int main(int argc, char *argv[])
{
   int a=0;
   double b[1000];

   do{
          cin>>b[a];
    b[a]=sqrt(b[a]);
    a++;}
   while (!cin.eof());


   a=a-2;

   while (a>=0)
   {
   cout<<setiosflags(ios::fixed)<<setprecision(4);
   cout<<b[a]<<endl;
   a--;
}

          system("PAUSE");
    return EXIT_SUCCESS;
    }

program z;
 var n,k:byte; f:longint;
     s:string[30];
 procedure factorial1(n,k:byte);
  var q:integer;
  begin
   q:=n;
   while q>=k do
    begin
     f:=f*q;
     q:=q-k;
    end;
  end;
  procedure factorial2(n,k:byte);
   var q:integer;
  begin
   q:=n;
   while q>=(n mod k) do
    begin
     f:=f*q;
     q:=q-k;
    end;
  end;
begin
 read (n,s);
 k:=length(s);
 f:=1;
 IF n mod k=0 then factorial1(n,k)
  else factorial2(n,k);
 writeln (f);
end.

program z;
 var n,k:byte; f:longint;
     s:string[30];
 procedure factorial1(n,k:byte);
  var q:integer;
  begin
   q:=n;
   while q>=k do
    begin
     f:=f*q;
     q:=q-k;
    end;
  end;
  procedure factorial2(n,k:byte);
   var q:integer;
  begin
   q:=n;
   while q>=(n mod k) do
    begin
     f:=f*q;
     q:=q-k;
    end;
  end;
begin
 read (n,s);
 k:=length(s);
 f:=1;
 IF n mod k=0 then factorial1(n,k)
  else factorial2(n,k);
 writeln (f);
end.

My best execution time is 0.015s, how can I improve the performance?

Thanks a lot!

What is the answer to
3
1
2
3

According to the question, the answer should be
1
3
isn't it? But why the correct answer is
2
1
2


What is exactly the sentence below mean?
"Your task is to choose a few of given numbers (1 ≤ few ≤ N) so that the sum of chosen numbers is multiple for N"

Is it already correct that the answer to the above question is {1, 2} where as the minimum set which its sum divisible by 3 is {3}. (I'm not English native)

Edited by author 07.02.2010 23:20

V zadache Pahom i ovrag u mena WA 2 test. Wse moi testi prohodat. Pomogite razobratsa!

To solve this problem we must use 64-bit variable. When i wrote in my code "printf("%lld\n",res)" i've got WA on test number 7. But when i wrote "cout<<res<<endl",i've got AC. My friend had the same problem. So what does it mean?

I have TLE on test 7. Why it can be?