January 1st 1600 is sunday, not monday!

Fucking participants!
This problem is the most stupid problem I have ever seen.
Why the problem is to output the calendar but not to find the day of
week? It is very simple to find but it is IMPOSSIBLE to output! The
sample does not give the answers to many questions.
Why I MUST GUESS, GUESS and GUESS again?
...
Diman_YES[UT], the leader of Universal Team

7 2 2000

mon......[.7]..14...21...28
tue...1....8...15...22...29
wed...2....9...16...23.....
thu...3...10...17...24.....
fri...4...11...18...25.....
sat...5...12...19...26.....
sun...6...13...20...27.....

btw, to turn ' ' into '.', one can use
 sed 's/ /./g'

I wrote the program on Python 3.4. The output for '30 1 2012' is
mon........2....9...16...23..[30]
tue........3...10...17...24...31
wed........4...11...18...25.....
thu........5...12...19...26.....
fri........6...13...20...27.....
sat........7...14...21...28.....
sun...1....8...15...22...29.....
It should be right, but I still got WA.
I think there is too few data about the output format in the problem statement. For example, I had to go to forum to get that there must be 4 spaces between first and second coulumn and no space in the end of the second line in this case.

To admins: please add some rules, which format of output should be, because this example from one month shows me nothing.
1) number of spaces must be 3 everywhere except 1,2,3,4,5,6,7,8,9?
2) we must count day of week using normal calender? Rules about it are full, or some items are omitted???
3) What is in test 10??
P.S.
Is this correct for: 7 1 2006

mon.......2....9...16...23...30
tue.......3...10...17...24...31
wed.......4...11...18...25
thu.......5...12...19...26
fri.......6...13...20...27
sat.....[.7]..14...21...28
sun...1...8...15...22...29

Or correct is:
mon........2....9...16...23...30
tue........3...10...17...24...31
wed........4...11...18...25
thu........5...12...19...26
fri........6...13...20...27
sat......[.7]..14...21...28
sun...1....8...15...22...29
?????
I mean number of spaces.

Edited by author 03.03.2007 11:41

The answer for "30 1 2012":

mon........2....9...16...23..[30]
tue........3...10...17...24...31
wed........4...11...18...25.....
thu........5...12...19...26.....
fri........6...13...20...27.....
sat........7...14...21...28.....
sun...1....8...15...22...29.....

between first and second column there are exactly 4(!) '.';

I have WA2, but I don't know why.
What a trick in test 2? Or can you give me some tricky tests?

Please, check this test(or checker) - my program passes all tests correctly(all 365 tests for non-leaps years and 366 for leap years), but WA 10, or say, if my solution gives a wrong answer at all :)

If I get WA#10 then I'm not right, but all tests in the forum I solved correcyly! It is so strnage! If you have some idea about this, may you write me? thank's a lot

What does the test WA#3? Help!!! And where does a leap year? : "Do not forget about the leap years."  Can you please show an example of a test WA#3. I write (programmed in) to C#. Thanks in advance!

Edited by author 03.03.2013 04:07

Edited by author 03.03.2013 16:50

[edit]  I used formula for day of week instead of pascal function. Got ac.
I am using dayofweek to get the weekday for the first of the month in the question.
I am getting WA7.
I think my formatting is OK but I am getting wrong answers for some inputs.
The two supplied examples are OK, and
20 10 2050
mon        3   10   17   24   31
tue        4   11   18   25
wed        5   12   19   26
thu        6   13  [20]  27
fri        7   14   21   28
sat   1    8   15   22   29
sun   2    9   16   23   30

and
25 11 1885
mon        2    9   16   23   30
tue        3   10   17   24
wed        4   11   18  [25]
thu        5   12   19   26
fri        6   13   20   27
sat        7   14   21   28
sun   1    8   15   22   29

but not
4 7 1890
mon      [ 4]  11   18   25
tue        5   12   19   26
wed        6   13   20   27
thu        7   14   21   28
fri   1    8   15   22   29
sat   2    9   16   23   30
sun   3   10   17   24   31

or
1 1 1600
mon [ 1]   8   15   22   29
tue   2    9   16   23   30
wed   3   10   17   24   31
thu   4   11   18   25
fri   5   12   19   26
sat   6   13   20   27
sun   7   14   21   28

from what I have seen in the forum.

Any ideas please?
Thanks




Edited by author 05.12.2012 22:33

test:
31 12 2400

be careful of the ']'

Subject

What is test 5? I can't find my mistake.

Edited by author 24.01.2011 16:15

Edited by author 24.01.2011 16:15

According to the discussion board.
I reference one code on here and change some thing.
But why still got the wrong answer.
I use the sample test to test.
The answer is right.
But when I upload to judgement.
Alway got the WA #1.

Someone can tell me why.
Thank you.

Below is my code.

#include<stdio.h>
#include<stdlib.h>

int is_leap_year(int y){
      return (  ( y%4==0 ) && (y%100!=0||y%400==0)  );
}


int days_per_year(int y){

      if(is_leap_year(y))
            return 366;
      else
            return 365;
}

int days_per_month(int m,int y){

      const int days[12]={31,28,31,30,31,30,31,31,30,31,30,31};
      if(m!=2)
             return days[m-1];
      else if(is_leap_year(y))
             return 29;
      else
             return 28;
}

int get_day_of_week(int m,int d,int y){
    int day=4;  /*  Key point  */
    int i;

      for(i=1600;i<y;i++) day+=days_per_year(i);
      for(i=1;i<m;i++) day+=days_per_month(i,y);
      day+=d;


return day%7;
}

void print_calendar(int m,int d,int y)
{
      const char* day_name[7]={"mon","tue","wed","thu","fri","sat","sun"};
      int days=days_per_month(m,y);
      int day_begin=1-get_day_of_week(m,1,y);

      int more_print = (day_begin + days)%7;

      int i,j,flag;


      flag = 0;
      for(i=0;i<7;i++){
          printf("%s",day_name[i]);

          for(j=day_begin+i ; j <= days + more_print ; j+=7 )

                if(  j<1   ||   j>days   )
                    printf("....");

                else if(j==d){
                        flag = 1;
                        if(j<10)
                           printf("..[%d]",j);
                        else
                           printf("..[%2d]",j);
                }
                else{
                              if(j<day_begin+7){
                                  if(flag){
                                     printf("..%d",j);
                                     flag = 0 ;
                                  }
                                  else{
                                    printf("...%d",j);
                                  }
                              }
                              else if(j >= day_begin+7 && j<10){
                                  if(flag){
                                     printf("...%d",j);
                                     flag = 0 ;
                                  }
                                  else{
                                    printf("....%d",j);
                                  }
                              }
                              else if(j>=10 && j<= days){
                                  if(flag){
                                     printf("..%2d",j);
                                     flag = 0 ;
                                  }
                                  else{
                                    printf("...%2d",j);
                                  }
                              }
                              else{
                                    printf("....");

                              }
                }/* if(  j<1   ||   j>days   )  */

                 printf("\n");
      }/* for(i=0;i<7;i++) */

}


int main()
{
    int d,m,y;
    scanf("%d %d %d",&d,&m,&y);

    if( y < 1600 || y > 2400) exit(0);
    if( m < 1 || m > 12 ) exit(0);
    if( d < 1 || d > 31 ) exit(0);

    print_calendar(m,d,y);


exit (0);
}

Some tests:

20 10 2050
mon........3...10...17...24...31
tue........4...11...18...25.....
wed........5...12...19...26.....
thu........6...13..[20]..27.....
fri........7...14...21...28.....
sat...1....8...15...22...29.....
sun...2....9...16...23...30.....
//////////////////////////////
4 7 1890
mon........7...14...21...28
tue...1....8...15...22...29
wed...2....9...16...23...30
thu...3...10...17...24...31
fri.[.4]..11...18...25.....
sat...5...12...19...26.....
sun...6...13...20...27.....
/////////////////////////////
'.' it's ' '

Good luck!

Edited by author 01.10.2009 06:11

Some tests:

20 10 2050
mon        3   10   17   24   31
tue        4   11   18   25
wed        5   12   19   26
thu        6   13  [20]  27
fri        7   14   21   28
sat   1    8   15   22   29
sun   2    9   16   23   30
//////////////////////////////
4 7 1890
mon        7   14   21   28
tue   1    8   15   22   29
wed   2    9   16   23   30
thu   3   10   17   24   31
fri [ 4]  11   18   25
sat   5   12   19   26
sun   6   13   20   27

Good luck!

I solved this problem using:
Calendar calendar = new GregorianCalendar(year,month-1,1)
It helped me to get position of the first day of month and day's count at month.

Pay attention to:
1)Day's count in month.
2)Use trim() to remove  ending spaces.

Here is my code :
{$M 16777216}
Program P1201_Which_day_is_it;
{$APPTYPE CONSOLE}
Var Days:array [1600..2400] of byte;
    fday,day,sum,month:byte;
    year:word;
    i,j,k,h,d,code:integer;
    s,t:string;
    Calendar:array[1..7] of string;
{-----------------------------------------------------------}
Procedure Init;
Begin
  Days[1600]:=2;
  i:=1601;
  While i<=year do
    Begin
      days[i]:=days[i-1];
      if (i-1) mod 4 =0 then inc(days[i]);
      if i-1=1918 then inc(days[i]);
      inc(days[i]);
      if days[i]>7 then days[i]:=days[i] mod 7;
      inc(i);
    end;
End;
{-----------------------------------------------------------}
Begin
 {$IFNDEF ONLINE_JUDGE}
 Assign(input,'Input.txt'); Reset(input);
 Assign(output,'Output.txt'); Rewrite(output);
 {$ENDIF}
 Readln(input,day,month,year);
 init; fday:=days[year];
 For i:=1 to month-1 do
   Begin
     if (i=1)or(i=3)or(i=5)or(i=7)or(i=8)or(i=10) then inc(fday,3);
     if (i=2) then
       if year mod 4 = 0 then inc(fday,1);
     if (i=4)or(i=6)or(i=9)or(i=11)or(i=12) then inc(fday,2);
   End;
 if (month=1)or(month=3)or(month=5)or(month=7)or(month=8)or(month=10) then
   sum:=31;
 if (month=2) then if (year mod 4=0) then sum:=29 else sum:=28;
 if (month=4)or(month=6)or(month=11)or(month=12) then sum:=30;
 Calendar[7]:='sun'; Calendar[1]:='mon'; Calendar[2]:='tue';
 Calendar[3]:='wed'; Calendar[4]:='thu'; Calendar[5]:='fri';
 Calendar[6]:='sat';
 if fday>7 then fday:=fday mod 7;
 for i:=fday-1 downto 1 do Calendar[i]:=Calendar[i]+'     ';
 k:=fday;
 for i:=1 to sum do
   begin
     if i=day then s:=']' else s:=' ';
     str(i,t); if length(t)=1 then t:=' '+t; s:=t+s;
     if i=day then s:='['+s;
     while length(s)<5 do s:=' '+s;
     Calendar[k]:=Calendar[k]+s;
     if k=7 then k:=1 else inc(k);
   end;
 For i:=1 to 7 do
   Begin
     While Calendar[i][length(calendar[i])]=' ' do
       delete(Calendar[i],Length(calendar[i]),1);
     writeln(output,Calendar[i]);
   End;
 {$IFNDEF ONLINE_JUDGE}
 Close(input);
 Close(output);
 {$ENDIF}
End.

My code have WA#6, but I think it alright !!

Was it Tuesday or Saturday???

Edited by author 23.10.2008 17:36