HDU 4965 fast matrix calculation matrix fast power

Link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4965

Question: a matrix of N * K, a matrix of K * n, B (4 <= n <=, 2 <= k <= 6 ). First, perform matrix multiplication c = a * B, then perform the power operation of D = C ^ (N * n), and then modulo 6 for each number of D, finally, the sum of all the positions and numbers of D is obtained.

Idea: Like the previous matrix multiplication, it is impossible to directly multiply a very large matrix without the help of small rules (currently, even the strassen matrix algorithm will not speed up to the required) the C matrix given in the question is a 1000*1000 matrix, and the rapid power is timed out. So I noticed that the number of columns in matrix A and the number of rows in matrix B cannot exceed six, however, it is no problem to perform a rapid power operation on the 6*6 matrix. D = a * B *... * a * B, which can be viewed as D = A * (B * *... * B * A) * B. In this way, the relationship between left multiplication and right multiplication of the matrix is not affected, and the number of rows and columns is too large. (The small details in the question are very powerful. First, the stack explosion problem needs to be solved, and then the matrix assignment operation is also the complexity of O (N)

Code:

# Pragma comment (linker, "/Stack: 1024000000,1024000000 ") # include <algorithm> # include <cmath> # include <cstdio> # include <cstdlib> # include <cstring> # include <ctime> # include <ctype. h> # include <iostream> # include <map> # include <queue> # include <set> # include <stack> # include <string> # include <vector> # define EPS 1e-8 # define INF 0x7fffffff # define PI ACOs (-1.0) # define seed 31 // 131,1313 typedef long ll; typedef unsigned long ull; # define mod 6 # define maxn 1005 # define maxm 1005 using namespace STD; struct matrix {int N, m; int A [maxn] [maxm]; void Init () {n = m = 0; memset (A, 0, sizeof ());} matrix Operator + (const Matrix & B) const {matrix TMP; TMP. N = N; TMP. M = m; For (INT I = 0; I <n; I ++) for (Int J = 0; j <m; j ++) TMP. A [I] [J] = A [I] [J] + B. A [I] [J]; return TMP;} Matrix Operator-(const Matrix & B) const {matrix TMP; TMP. N = N; TMP. M = m; For (INT I = 0; I <n; I ++) for (Int J = 0; j <m; j ++) TMP. A [I] [J] = A [I] [J]-B. A [I] [J]; return TMP;} Matrix Operator * (const Matrix & B) const {matrix TMP; TMP. N = N; TMP. M = B. m; For (INT I = 0; I <n; I ++) {for (Int J = 0; j <B. m; j ++) TMP. A [I] [J] = 0 ;}for (INT I = 0; I <n; I ++) for (Int J = 0; j <B. m; j ++) for (int K = 0; k <m; k ++) {TMP. A [I] [J] + = A [I] [k] * B. A [k] [J] ;}for (INT I = 0; I <n; I ++) for (Int J = 0; j <B. m; j ++) TMP. A [I] [J] % = MOD; return TMP;} void copy (const Matrix & X) {n = x. n; M = x. m; For (INT I = 0; I <n; I ++) for (Int J = 0; j <m; j ++) A [I] [J] = x. A [I] [J] ;}}; // A × B makes sense only when the number of columns in matrix A is equal to the number of rows in matrix B. Matrix m_quick_pow (matrix m, int K) {matrix TMP; TMP. N = m. n; TMP. M = m. m; // M = N can be used for Fast Power (INT I = 0; I <TMP. n; I ++) {for (Int J = 0; j <TMP. n; j ++) {if (I = J) TMP. A [I] [J] = 1; else TMP. A [I] [J] = 0 ;}while (k) {If (K & 1) TMP. copy (TMP * m); k> = 1; M. copy (M * m);} return TMP;} matrix A, B; int main () {int N, K; while (scanf ("% d ", & N, & K) {If (n = 0 & K = 0) break;. N = N;. M = K; For (INT I = 0; I <. n; I ++) for (Int J = 0; j <. m; j ++) scanf ("% d", &. A [I] [J]); B. N = K; B. M = N; For (INT I = 0; I <B. n; I ++) for (Int J = 0; j <B. m; j ++) scanf ("% d", & B. A [I] [J]); matrix E, C; C. copy (B * A); C. copy (m_quick_pow (C, N * N-1); E. copy (A * C); E. copy (E * B); LL ans = 0; For (INT I = 0; I <E. n; I ++) {for (Int J = 0; j <E. m; j ++) {ans + = (LL) (E. A [I] [J]) ;}} printf ("% i64d \ n", ANS);} return 0 ;}