HDu 1121 Complete the Sequence(DP多項式差分)

Complete the Sequence
Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 161    Accepted Submission(s): 93
Problem Description
You probably know those quizzes in Sunday magazines: given the sequence 1, 2, 3, 4, 5, what is the next number? Sometimes it is very easy to answer, sometimes it could be pretty hard. Because these "sequence problems"are very popular, ACM wants to implement them into the "Free Time"section of their new WAP portal.
ACM programmers have noticed that some of the quizzes can be solved by describing the sequence by polynomials. For example, the sequence 1, 2, 3, 4, 5 can be easily understood as a trivial polynomial. The next number is 6. But even more complex sequences, like 1, 2, 4, 7, 11, can be described by a polynomial. In this case, 1/2.n^2-1/2.n+1 can be used. Note that even if the members of the sequence are integers, polynomial coefficients may be any real numbers.
Polynomial is an expression in the following form:
P(n) = aD.n^D+aD-1.n^D-1+...+a1.n+a0
. If aD <> 0, the number D is called a degree of the polynomial. Note that constant function P(n) = C can be considered as polynomial of degree 0, and the zero function P(n) = 0 is usually defined to have degree -1.
 
Input
There is a single positive integer T on the first line of input. It stands for the number of test cases to follow. Each test case consists of two lines. First line of each test case contains two integer numbers S and C separated by a single space, 1 <= S < 100, 1 <= C < 100, (S+C) <= 100. The first number, S, stands for the length of the given sequence, the second number, C is the amount of numbers you are to find to complete the sequence.
The second line of each test case contains S integer numbers X1, X2, ... XS separated by a space. These numbers form the given sequence. The sequence can always be described by a polynomial P(n) such that for every i, Xi = P(i). Among these polynomials, we can find the polynomial Pmin with the lowest possible degree. This polynomial should be used for completing the sequence.
 
Output
For every test case, your program must print a single line containing C integer numbers, separated by a space. These numbers are the values completing the sequence according to the polynomial of the lowest possible degree. In other words, you are to print values Pmin(S+1), Pmin(S+2), .... Pmin(S+C).
It is guaranteed that the results Pmin(S+i) will be non-negative and will fit into the standard integer type.
 
Sample Input

   
   
   
   

    
    
    
    4
6 3
1 2 3 4 5 6
8 2
1 2 4 7 11 16 22 29
10 2
1 1 1 1 1 1 1 1 1 2
1 10
3
   
   
   
   

 
Sample Output

   
   
   
   

    
    
    
    7 8 9
37 46
11 56
3 3 3 3 3 3 3 3 3 3
   
   
   
   

 
Source
Central Europe 2000
 
Recommend
JGShining
 
考え方:1つの数列の前のいくつかの項目は、後の項目を推測することを要求します.ラグランジュ補間法は容易に考えられるが,精度は大きな問題となっている.
補間法に加えて,この数列問題を解くにはより良い差分法があり,過程で浮動小数点数操作には全く関与しない.例えば、1 2 4 7 11 16 22 29という数列について、私たちは各項目について前の項目との差、すなわち2-1=1、4-2=2、7-4=3、....このようにして、1次差分:1、2、3、4、5、6、7を得る.さらに2次差分を求めたのは,1,1,1,1,1,1であった.この時、法則はすでに少し明らかになった.
すなわち,任意の合理的多項式通項が存在する数列に対して差分法で可能である.
したがって,このn項数列のn−1次差分を求めて逆算すればよい.次は簡単です.
 

#include<iostream>
#include<cstring>
using namespace std;
const int mm=110;
int f[mm][mm];
int main()
{
  int cas;
  while(cin>>cas)
  {
    while(cas--)
    {
      int m,n;
      cin>>m>>n;
      for(int i=0;i<m;i++)cin>>f[0][i];
      for(int i=1;i<m;i++)
        for(int j=0;j+i<m;j++)
        f[i][j]=f[i-1][j+1]-f[i-1][j];
      for(int i=1;i<=n;i++)///    n  ,  m+n         m 
        f[m-1][i]=f[m-1][0];
      for(int i=m-2;i>=0;i--)///  
      for(int j=0;j<n;j++)
        f[i][m-i+j]=f[i+1][m-i+j-1]+f[i][m-i+j-1];
        cout<<f[0][m];
      for(int i=1;i<n;i++)
        cout<<" "<<f[0][m+i];
      cout<<"
";
    }
  }
}