HDU 4551 birthday prediction (simple question)

Birthday Prediction

Time Limit: 3000/1000 MS (Java/others) memory limit: 65535/32768 K (Java/Others)
Total submission (s): 460 accepted submission (s): 176

Problem description James is very important to his birthday, because he can get a blessing, share happiness with his friends and family, and make a summary of his life, and he can receive many gifts!
James is a mysterious person and won't tell you his birthday easily. Now he has a way to guess the day of his birthday.

James will tell you the following information:

1. the maximum number of common appointments between the month of birth and the day of birth;
2. Minimum Public multiples of the birth month and date;
3. Year of birth;

I want you to guess James's birthday.

 

In the first line of input, enter a positive integer T, indicating that there are a total of T groups of data (T <= 200 );
For each group of data, input three numbers X, Y, Z,
X indicates the maximum number of common appointments between the month of birth and the day of birth (1 <= x <= 1000 );
Y indicates the minimum public multiple of the birth month and date (1 <= Y <= 1000 );
Z indicates the Year of birth (1900 <= z <= 2013 ).
Each group of input data occupies one row.

 

Output first outputs the number of cases for each group of data.
If the answer does not exist, output "-1 ";
If the answer exists but is not unique, "1" is output ";
If the answer is unique, the output date is in the format of yyyy/mm/DD;
Each group of outputs occupies one row. For the specific output format, see the example.

 

Sample input3
12 24 1992
3 70 1999
9 18 1999

 

Sample outputcase #1: 1992/12/24
Case #2:-1
Case #3:

 

Source2013 Jinshan xishanju creative game program challenge-Preliminary Round (3)

 

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 // ========================================================== ==========================================  //  Name: A. cpp  //  Author:  //  Version:  //  Copyright: Your copyright notice  //  Description: Hello world in C ++, ANSI-style  //  ========================================================== ========================================== # Include <Iostream> # Include <Stdio. h> # Include < String . H> # Include <Algorithm> # Include <Map> # Include <Queue> # Include < Set > # Include <Vector> # Include < String ># Include <Math. h> Using   Namespace  STD;  Int Gcd ( Int A, Int  B ){  If (B = 0 ) Return  A;  Return Gcd (B, A % B );}  Int Day [] = {0 , 31 , 28 , 31 , 30 , 31 , 30 , 31 , 31 , 30 , 31 , 30 , 31  };  Bool Check ( Int Y, Int M, Int  D ){  If (M> 12 ) Return   False  ;  If (M = 2  ){  If (Y % 400 = 0 | (Y %100 ! = 0 & Y % 4 = 0  )){  If (D> 29 ) Return   False  ;}  Else  {  If (D> 28 ) Return  False  ;}}  Else  {  If (D> day [m]) Return   False  ;}  Return   True  ;}  Int  Main (){  //  Freopen ("in.txt", "r", stdin );  // Freopen ("out.txt", "W", stdout );      Int  X, Y, Z;  Int  T; scanf (  "  % D  " ,& T );  Int Icase = 0  ;  While (T -- ) {Icase ++ ; Scanf ( "  % D  " , & X, & Y ,& Z); printf (  "  Case # % d:  "  , Icase );  Int TMP = x * Y;  Int Ans = 0  ;  Int  Ansm, anSd;  For ( Int I = 1 ; I <= 12 ; I ++ ){  If (TMP % I) Continue  ;  Int Temp = tmp/ I;  If (Gcd (I, temp) = x && Check (Z, I, temp) {ans ++ ; Ansm =I; anSd = Temp ;}}  If (ANS = 0 ) Printf ( "  -1 \ n  "  );  Else   If (ANS> 1 ) Printf ( "  1 \ n  "  ); Else Printf ( "  % D/% 02d/% 02d \ n  "  , Z, ansm, anSd );}  Return   0  ;}