HDU 5015 233 matrix (constructor matrix)

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 5015

Because it is a two-dimensional recursive formula, we didn't expect to construct a matrix like this. From the column, the current column is obtained by recursion of the previous column. Construct a matrix based on this point. B [I] represents column I, which is a matrix (n + 2) * 1, that is, B [1] = [1,233 ......], the reason why two rows are added is to obtain the second number 2333 in B [I] from a matrix B [I-1 ..., construct a conversion matrix A, which is a matrix (n + 2) * (N + 2). Then a ^ m-1) * B is the m column.

/* A matrix: 1 0 0 0 0... 3 10 0 0 0... 3 10 1 0 0... 3 10 1 1 0... 3 10 1 1 1 ..... matrix B: 1st columns */# include <stdio. h> # include <iostream> # include <map> # include <set> # include <list> # include <stack> # include <vector> # include <math. h> # include <string. h> # include <queue> # include <string> # include <stdlib. h >#include <algorithm> # define ll _ int64 // # define ll long # define EPS 1e-9 # define PI ACOs (-1.0) using namespace STD; Co NST int INF = 0x3f3f3f; const int mod = 10000007; struct matrix {ll mat [15] [15]; void Init () {memset (MAT, 0, sizeof (MAT); For (INT I = 0; I <15; I ++) {mat [I] [I] = 1 ;}} A, B, res; int n, m; matrix MUL (matrix A, matrix B) {matrix ans; memset (ans. mat, 0, sizeof (ans. mat); For (INT I = 0; I <n + 2; I ++) {for (int K = 0; k <n + 2; k ++) {if (. mat [I] [k] = 0) continue; For (Int J = 0; j <n + 2; j ++) {ans. mat [I] [J] + = (. mat [I] [k] * B. mat [k] [J]) % MOD; ans. mat [I] [J] % = mod ;}} return ans;} matrix POW (matrix A, int N) {matrix ans; ans. init (); While (n) {If (N & 1) ans = MUL (ANS, a); N >>= 1; A = MUL (a, );} return ans;} int main () {int X; while (~ Scanf ("% d", & N, & M) {memset (B. mat, 0, sizeof (B. mat); B. mat [0] [0] = 1; B. mat [1] [0] = 233; For (INT I = 2; I <n + 2; I ++) {scanf ("% d", & X ); b. mat [I] [0] = (B. mat [I-1] [0] + x % mod) % MOD;} memset (. mat, 0, sizeof (. mat);. mat [0] [0] = 1; for (INT I = 1; I <n + 2; I ++) {. mat [I] [0] = 3;. mat [I] [1] = 10; for (Int J = 2; j <= I; j ++). mat [I] [J] = 1;} res = POW (A m-1); ll anw = 0; For (INT I = 0; I <n + 2; I ++) {ANW + = (res. mat [n + 1] [I] * B. mat [I] [0]) % MOD; ANW % = MOD;} printf ("% i64d \ n", ANW);} return 0 ;}

HDU 5015 233 matrix (constructor matrix)