Poj 3150 cellular automaton (matrix multiplication + binary)

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The number of N is given to form a ring. One transformation is to take the remainder of the d number before and after each number, and then use it as the number. After K conversions, what is an array.

Train of Thought: here

First, let's take a look at the first group of data in the sample.
1 2 2 1 2
After a transformation
5 5 5 5 4
The principle is
A0 A1 A2 A3 A4
->
(A4 + A0 + A1) (A0 + A1 + A2) (A1 + A2 + A3) (A2 + A3 + A4) (A3 + A4 + a0)

If we use a matrix multiplication to describe it, we can express it as a matrix multiplication of 1xn and nxn, and the result is still 1xn matrix.
A = 1 2 2 1 2
B =
1 1 0 0 1
1 1 1 0 0
0 1 1 1 0
0 0 1 1 1
1 0 0 1 1
A * B = 5 5 5 4
So the final result is: A * (B ^ K)

When the linear algebra is unqualified, the shoes indicate great pressure ..

Calculate the k power of an nxn matrix, and the K is large, and N is not small. What should I do?
As a result, some experts have observed that this matrix is somewhat special and can find some rules:
B ^ 1 =
[1, 1, 0, 0, 1]
[1, 1, 1, 0, 0]
[0, 1, 1, 1, 0]
[0, 0, 1, 1, 1]
[1, 0, 0, 1, 1]
B ^ 2 =
[3, 2, 1, 1, 2]
[2, 3, 2, 1, 1]
[1, 2, 3, 2, 1]
[1, 1, 2, 3, 2]
[2, 1, 1, 2, 3]
B ^ 3 =
[7, 6, 4, 4, 6]
[6, 7, 6, 4, 4]
[4, 6, 7, 6, 4]
[4, 4, 6, 7, 6]
[6, 4, 4, 6, 7]
B ^ 4 =
[19, 17, 14, 14, 17]
[17, 19, 17, 14, 14]
[14, 17, 19, 17, 14]
[14, 14, 17, 19, 17]
[17, 14, 14, 17, 19]

No. Is whether it is the several power of B, are in line with a [I] [J] = A [I-1] [J-1]
The experts say this is how it is pushed down:
""
Using Matrix A, B has a [I] [J] = A [I-1] [J-1], B [I] [J] = B [I-1] [J-1] (I-1 <0 represents I-1 + N, J-1 <0 represents J-1 + n)
We can conclude that the matrix C = a * B also has this property.
C [I] [J] = sum (A [I] [T] * B [T] [J]) = sum (A [I-1] [T-1], B [T-1] [J-1]) = sum (A [I-1] [T], B [T] [J-1]) = C [I-1] [J-1]
"

In this way, an array of N sizes can be opened to store the results of each calculation. Nxn is not necessary.
N has been solved, but K is still very large. What should I do?

In this case, we can use the bipartite method to obtain B ^ K.
B ^ K = B ^ 1 * B ^ 4 * B ^ 16...

1 // 3150 2 # include <cstdio> 3 # include <cstring> 4 # include <iostream> 5 # define ll long 6 7 using namespace STD; 8 9 int N, m, D, K; 10 ll a [1010], B [1010]; 11 void multi (LL * C, ll * D) 12 {13 LL X [1010]; 14 For (INT I = 0; I <n; I ++) 15 {16 x [I] = 0; 17 for (Int J = 0; j <N; j ++) 18 x [I] + = C [J] * d [I> = J? (I-j): (n + I-j)]; // prevents negative numbers, forming a ring 19} 20 for (INT I = 0; I <N; I ++) 21 d [I] = x [I] % m; 22} 23 int main () 24 {25 while (CIN> N> m> D> K) {26 for (INT I = 0; I <n; I ++) 27 Cin> A [I]; 28 B [0] = 1; 29 for (INT I = 1; I <= D; I ++) 30 B [I] = B [n-I] = 1; 31 While (k) 32 {33 If (K & 1) // odd 34 multi (B, ); 35 multi (B, B); 36 K >>= 1; 37} 38 for (INT I = 0; I <n; I ++) 39 if (I = N-1) printf ("% i64d \ n", a [I]); 40 else printf ("% i64d", a [I]); 41} 42 return 0; 43}

View code

 

In the calculation process, the number must be greater than M.
Remember x * Y = (X % m) * (Y % m)