CMPLS - Complete the Sequence!

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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.n2-1/2.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.nD+aD-1.nD-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 (equal to about 5000). 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.

Example

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
Warning: large Input/Output data, be careful with certain languages

hide comments
flyingduchman_: 2016-11-19 19:52:36

Use difference method.
Algo: given array a of size n, on first step make an array a1 of size n-1 by the keeping the difference of two adjacent element of given array a. Like:a1[0] = a[1]-a[0] .....
a1[n-2] = a[n-1]-a[n-2]. Say array a1 as first row of difference table. Keep making rows until all elements in a row are same or there are only one element in a row.
so make a2, a3 .... until all elements become same or have only one element left. Now it's time to stop and expand those rows in a bottom-up manner. Let you made a table of size k = 3, a1(n-1 elements), a2(n-2 elements), a3(n-3 elements).
Now expand the last row of the difference table by "c" elements, with same elments or the remaining one.Here a3 will be expanded with "c" elements. Now you can expand the last element of a2[n-1](new expanded element) = a2[n-2]+a3[n-2](extended by constant difference), here array index starts from 0.
similarly, a1[n-1] = a1[n-2]+a2[n-3]
now original array, a[n] = a[n-1]+a1[n-2] is expanded.
If you repeat the bottom-up procedure "c" times, original array will be expanded by c number of elements. You can use "std vector" in c++

Last edit: 2016-11-19 19:56:49
mredik5: 2016-07-20 15:07:49

For me this problem is about finding the formula of polynomial Pmin that gives that output not to find . So basically you shouldnt care about the 3rd case just do a program that will give you the right answer. I can help some of you that the lowest possible digree of Pmin should not(i think it wont) be higher than C.
the 3rd case's formula is:
a(n) = (1/9!)*(n+1)*(n^8 - 37*n^7 + 583*n^6 - 5119*n^5 + 27568*n^4 - 94852*n^3 + 212976*n^2 - 322560*n + 9!) (based on http://oeis.org/A115205)

get_right_jr: 2016-07-04 09:59:31

Easy problem
Use Difference Method (make Forward Difference Table)
0.1s in Java-AC in one go! :)

more_practice: 2016-06-27 12:12:56

Try difference of method; in RECursion,it is of 3.13 sec and in iteration 0.05sec.

Parikshit: 2016-06-23 12:06:23

simple method of difference AC in one go.. 0.04 sec

Last edit: 2016-06-23 12:17:39
Sarthak Munshi: 2016-06-08 07:52:35

strange ! cannot figure out polynomial finite difference theorem for 3rd test case !

salvatore13: 2016-04-18 23:07:12

i don't know but wolfram alpha knows in some magic way
https://www.wolframalpha.com/input/?i=1+1+1+1+1+1+1+1+1+2+11+56

ekranoplano: 2016-04-08 23:38:47

Guys, It is possible to have some new test cases? Tnx

nathanek: 2016-02-25 13:20:39

guys im lost, i have used finite difference method and am getting the wrong answer. Any tips on test cases?

layman806: 2016-01-23 19:53:26

@enigmus , I used method of difference too, and I got it right in 0.15s on C++ 5.1 !


Added by:adrian
Date:2004-05-08
Time limit:5s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: NODEJS PERL6 VB.NET
Resource:ACM Central European Programming Contest, Prague 2000