Nyoj 1075 (recursive + rapid matrix power)

Red virus problem time limit: 1000 MS | memory limit: 65535 kb difficulty: 4

Description

Medical researchers have recently discovered a new virus, known as the red virus, because it spreads at a speed comparable to the "red code" that once spread over the Internet ". It was found that, in the DNA sequence of the virus and its variants, both of them are paired. Lyh wants to know the number of all DNA sequences that may become the virus under this feature.

Input

Multiple groups of test data. Enter an integer N for each data set to indicate the length of the virus DNA sequence. (1 ≤ n ≤ 10 ^ 9) If n = 0, the input is over and no processing is required.

Output

Each group of output occupies one row, representing the number of all possible DNA sequences with the virus length of N. Because the result may be very large, you only need to output the result after the remainder of 10007.

Sample Input

12

Sample output

26

Prompt

The DNA sequence is composed of four types of nucleotide, namely, "A", "g", "C", and "T. When n = 2, all possible DNA sequences are TT, TG, GT, GG, AA, and CC.

Analysis: because the question requires that a and c appear in pairs, that is, the numbers of A and C are both even, so four states can be defined:

DP [N] [0] indicates that when the length is N, the number of A is an even number, and the number of C is also an even number,

DP [N] [1] indicates that when the length is N, the number of A is an even number, and the number of C is an odd number,

DP [N] [2] indicates that when the length is N, the number of A is an odd number, and the number of C is an even number,

DP [N] [3] indicates that when the length is N, the number of A is an odd number, and the number of C is also an odd number of DNA sequences,

Then

DP [N] [0] = DP [n-1] [0] * 2 + dp [n-1] [1] * 1 + dp [n-1] [2] * 1 + dp [n-1] [3] * 0

DP [N] [1] = DP [n-1] [0] * 1 + dp [n-1] [1] * 2 + dp [n-1] [2] * 0 + dp [n-1] [3] * 1

DP [N] [2] = DP [n-1] [0] * 1 + dp [n-1] [1] * 0 + dp [n-1] [2] * 2 + dp [n-1] [3] * 1

DP [N] [3] = DP [n-1] [0] * 0 + dp [n-1] [1] * 1 + dp [n-1] [2] * 1 + dp [n-1] [3] * 2

Based on this recursive relationship, a matrix such as the next 4*4 can be constructed,

| 2 1 1 0 |

| 1 2 0 1 |

| 1 0 2 1 |

| 0 1 1 2 |

Then, the answer can be quickly obtained using the matrix's quick power.

# Include <cstdio> # include <cstring> # define mod 10007 struct matrix {int mat [4] [4]; matrix () {memset (MAT, 0, sizeof (MAT); For (INT I = 0; I <4; I ++) mat [I] [I] = 1 ;}}; matrix multi (matrix, matrix B) {matrix res; For (INT I = 0; I <4; I ++) {for (Int J = 0; j <4; j ++) {res. mat [I] [J] = 0; For (int K = 0; k <4; k ++) {res. mat [I] [J] + =. mat [I] [k] * B. mat [k] [J]; Res. mat [I] [J] % = mod ;}} return res;} matrix POW (matrix A, int X) {matrix res; while (X) {If (X & 1) RES = multi (Res, a); A = multi (a, a); X >>=1;} return res ;} int main () {int t, n; matrix A; // coefficient matrix. mat [0] [0] = 2;. mat [0] [1] = 1;. mat [0] [2] = 1;. mat [0] [3] = 0;. mat [1] [0] = 1;. mat [1] [1] = 2;. mat [1] [2] = 0;. mat [1] [3] = 1;. mat [2] [0] = 1;. mat [2] [1] = 0;. mat [2] [2] = 2;. mat [2] [3] = 1;. mat [3] [0] = 0;. mat [3] [1] = 1;. mat [3] [2] = 1;. mat [3] [3] = 2; scanf ("% d", & T); While (t --) {scanf ("% d", & N ); matrix ans = POW (A, n); printf ("% d \ n", ans. mat [0] [0]);} return 0 ;}

Nyoj 1075 (recursive + rapid matrix power)