Hdu1121 complete the sequence

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1121

Here is a series of integers that contain S integers, so that you can find the polynomial P (n) = AD. N ^ d + aD-1.n ^ D-1 +... + a1.n + a0: When the degree of the polynomial is the hour, the following C integers are output to satisfy the polynomial.

AlgorithmAnalysis: HDU recommended for DP, no idea, find the problem report (http://rchardx.is-programmer.com/posts/16142.html), said to be differential. My understanding is as follows: Find the difference between each item and the previous one, and then observe the law, for any polynomial satisfying P (n) = AD. N ^ d + aD-1.n ^ D-1 +... + a1.n + a0 of the d degree polynomial, take the first order of the difference: TMP = P (n)-(N-1) must be a D-1 degree polynomial, and so on, take the n-1 order difference, and there is only one number D (ProgramFor f [n-1] [0]), if D = 0, if you want to minimize the order of P (n), the next M number in the n-1 difference should be 0, if D! = 0, when the next M number is d, then the first order of the n-2 is a polynomial of the First Order (only the first degree of Polynomial (a1.n + a0, tolerances for A1) the difference is a constant), The n-1 order is 0 order polynomial, in order to ensure the minimum order d

 

Code

  # Include <  Stdio. h  >  
  # Define  Nn 102  
  Int  Main ()
{
  Int  T, I, J, S, C;
  Int  F [NN] [NN];
Scanf (  "  % D  "  , &  T );
  While  (T  --  ){
Scanf (  "  % D  "  ,  &  S,  &  C );
  For  (I  =    0  ; I  <  S; I  ++  )
Scanf (  "  % D  "  ,  &  F [  0  ] [I]);
  For  (I  =    1  ; I  <  S; I  ++  )
  For  (J  =     0  ; I  +  J  <  S; j  ++  )
F [I] [J] =  F [I  -     1  ] [J  +     1  ]  -  F [I  -     1  ] [J];
  For  (I =     1  ; I  <=  C; I  ++  )
F [s  -     1  ] [I]  =  F [s  -     1  ] [ 0  ];
  For  (I  =  S  -     2  ; I  > =     0  ; I  --  )
  For  (J =  S  -  I; j  <  C  +  S; j  ++  )
F [I] [J]  =  F [I] [J  -     1  ]  +  F [I +     1  ] [J  -     1  ];
  For  (I  =  S; I  <  S  +  C  -    1  ; I  ++  )
Printf (  "  % D  "  , F [  0  ] [I]);
Printf (  "  % D \ n  "  , F [  0  ] [I]);
}
  Return    0  ;
}