POJ 3233 Matrix Power Series（矩阵快速幂）

题意：

求 S = A + A² + A³ + … + A^k

思路：

和求矩阵幂方法类似，采取二分的思想：

A+A^2+A...+A^(2k+1)= A+A^2+...+A^k + A^(k+1) + A^(k+1)*(A+A^2+...+A^k).
A+A^2+...+A^2k = A+A^2+...+A^k + A^k*(A+A^2+...+A^k).

#include <cstdio> #include <cstring> #include <cstring> #include <iostream>

using namespace std; const int MAXN = 32; struct Matrix { int v[MAXN][MAXN]; } ; int n, M; Matrix matrixadd(Matrix a, Matrix b) { Matrix c; for (int i = 0; i < n; ++i) for (int j = 0; j < n; ++j) c.v[i][j] = (a.v[i][j] + b.v[i][j]) % M; return c; } Matrix matrixmul(Matrix a, Matrix b) { Matrix c; for (int i = 0; i < n; ++i) for (int j = 0; j < n; ++j) { c.v[i][j] = 0; for (int k = 0; k < n; ++k) c.v[i][j] = (a.v[i][k] * b.v[k][j] + c.v[i][j]) % M; } return c; } Matrix matrixpow(Matrix m, int k) { if (k == 1) return m; Matrix c; c = matrixpow(m, k >> 1); c = matrixmul(c, c); if (k % 2) c = matrixmul(c, m); return c; } Matrix matrixcalc(Matrix m, int k) { if (k == 1) return m; Matrix a, b, c; a = matrixcalc(m, k >> 1); if (k % 2) { b = matrixpow(m, (k+1)>>1); c = matrixadd(matrixadd(a, b), matrixmul(a, b)); } else { b = matrixpow(m, k >> 1); c = matrixadd(a, matrixmul(a, b)); } return c; } int main() { Matrix m; int k; scanf("%d %d %d", &n, &k, &M); memset(m.v, 0, sizeof(m.v)); for (int i = 0; i < n; ++i) for (int j = 0; j < n; ++j) scanf("%d", &m.v[i][j]); m = matrixcalc(m, k); for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) printf("%d ", m.v[i][j]); printf("\n"); } return 0; }