HDU children's queue

Children's queue

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)Total submission (s): 160 accepted submission (s): 102 

Problem description There are too students in PHT School. one day, the headmaster whose name is pigheader wanted all students stand in a line. he prescribed that girl can not be in single. in other words, either no girl in the queue or more than one girl stands side by side. the case n = 4 (n is the number of children) is like
FFFF, fffm, mfff, ffmm, mffm, mmff, Mmmm
Here F stands for a girl and M stands for a boy. the total number of queue satisfied the headmaster's needs is 7. can you make a program to find the total number of queue with N children?  

 

Input There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer N means the number of children (1 <=n <= 1000)  

 

Output For each test case, there is only one integer means the number of queue satisfied the headmaster's needs.  

 

Sample Input

 
123
 

  Sample output

 
124
 

  Note: 1. Recursive Formula 1) m

2) MFF

3) mfff

A: After the security sequence is followed by FF or m, the result is still safe.

B: Adding FF after the unsafe sequence can ensure its security. Although MF and F can also obtain the safe sequence, it is repeated in case.

Therefore: formula A [n] = A [n-1] + A [N-2] + A [n-4];

2. Large numbers and high precision. You can use a two-dimensional array to simulate large numbers. Output result when n = 1000:

12748494904808148294446671041721884239818005733501580815621713101333980596197474

74433619974245291299822523591089179822154130383839594330018972951428262366519975

47955743099808702532134666561848656816661065088789701201682837073071502397487823

19037

 # Include <stdio. h>  Int A [ 1001  ] [  101  ] = {  0  };  Void Add (  Int N ){  Int K =  0  , J;  For  ( J =  1  ; J <  101  ; J ++ ){ K + = A [ N - 1  ] [ J ] + A [ N -  2  ] [ J ] + A [ N -  4  ] [ J ]; A[ N ] [ J ] = K %  10000  ; K = K /  10000  ;  // Printf ("% d", k );  }  While ( K ){ A [ N ] [ J ++] = K %  10000  ; K = K /  10000  ;}} Int  Main  (){ A [  1  ] [  1  ] =  1  ; A [  2  ] [  1 ] =  2  ; A [  3  ] [  1  ] =  4  ; A [  4  ] [  1  ] = 7  ;  Int N , I ;  For  ( I =  5  ; I <  1001  ; I++ ){ Add ( I );}  While  ( Scanf (  "% D"  ,& N )! = EOF ){  For  ( I=  100  ; I >  0  ; I --){  If  ( A [ N ] [ I ]! =  0 )  Break  ;} Printf (  "% D"  , A [ N ] [ I ]);  For  ( I = I-  1  ; I >  0  ; I --){ Printf (  "% 04d"  , A [ N ] [ I ]);} Printf (  "\ N"  );}  Return  0  ;}