例题22

不错的矩阵乘法入门题
构造d*d矩阵

A=⎡⎣⎢⎢⎢⎢⎢⎢a1100a2010a3000a400...1............ad000⎤⎦⎥⎥⎥⎥⎥⎥

Fn−1=⎡⎣⎢⎢⎢⎢⎢⎢f(n−1)f(n−2)f(n−3)...f(n−d)⎤⎦⎥⎥⎥⎥⎥⎥

Fn=⎡⎣⎢⎢⎢⎢⎢⎢f(n)f(n−1)f(n−2)...f(n−d+1)⎤⎦⎥⎥⎥⎥⎥⎥

Fn=A×Fn−1
Fn=An−d×Fd

Fd=⎡⎣⎢⎢⎢⎢⎢⎢f(d)f(d−1)f(d−2)...f(1)⎤⎦⎥⎥⎥⎥⎥⎥
矩阵乘法不满足交换律，注意顺序！！！注意不要乘反！！！！！

#include<iostream>#include<string>#include<cstring>using namespace std;const int maxn = 20;typedef long long Matrix[maxn][maxn];typedef long long Vector[maxn];int sz, mod;void matrix_mul(Matrix A, Matrix B, Matrix res) {  Matrix C;  memset(C, 0, sizeof(C));  for(int i = 0; i < sz; i++)    for(int j = 0; j < sz; j++)      for(int k = 0; k < sz; k++) C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod;  memcpy(res, C, sizeof(C));}void matrix_pow(Matrix A, int n, Matrix res) {  Matrix a, r;  memcpy(a, A, sizeof(a));  memset(r, 0, sizeof(r));  for(int i = 0; i < sz; i++) r[i][i] = 1;  while(n) {    if(n&1) matrix_mul(r, a, r);    n >>= 1;    matrix_mul(a, a, a);  }  memcpy(res, r, sizeof(r));}void transform(Vector d, Matrix A, Vector res) {  Vector r;  memset(r, 0, sizeof(r));  for(int i = 0; i < sz; i++)    for(int j = 0; j < sz; j++)      r[j] = (r[j] + d[i] * A[i][j]) % mod;  memcpy(res, r, sizeof(r));}int main() {  int d, n, m;  while(cin >> d >> n >> m && d) {    Matrix A;    Vector a, f;    for(int i = 0; i < d; i++) { cin >> a[i]; a[i] %= m; }    for(int i = d-1; i >= 0; i--) { cin >> f[i]; f[i] %= m; }    memset(A, 0, sizeof(A));    for(int i = 0; i < d; i++) A[i][0] = a[i];    for(int i = 1; i < d; i++) A[i-1][i] = 1;    sz = d;    mod = m;    matrix_pow(A, n-d, A);    transform(f, A, f);    cout << f[0] << endl;  }  return 0;}