矩阵快速幂好题。题意:给一个N×K的矩阵A和K×N的矩阵B(4 <= N <= 1000,2 <= K <= 6),算出N×N的矩阵C=A×B,再算出N×N的矩阵M = C^(n×n),最后输出矩阵M每一个元素模6的和。
我的解题思路:矩阵C是N×N的矩阵,而N的值高达1000,因此暴力计算会超时。根据C=A×B,所以C^(n×n) = (A×B)^(n×n)。矩阵乘法满足乘法分配律,所以(A×B)^(n×n) = A×(B×A)^(n×n-1)×B,其中B×A是一个M×M的矩阵,M最多为6,这样就可以避免超时了。
我的解题代码:
#include <cstdio> #include <cstdlib> #include <cstring> #include <ctime> #include <climits> #include <algorithm> using namespace std; const int N = 1000; const int M = 6; const int MOD = 6; struct matrix { int n, m; int mat[M][M]; matrix(): n(0), m(0) {} matrix(int nn, int mm): n(nn), m(mm) {} void Unit(int unit) { n = m = unit; for (int i=0; i<n; ++i) for (int j=0; j<n; ++j) mat[i][j] = i == j ? 1 : 0; } }; int a[N][M], b[M][N], res[N][N]; int n, m; void InitRead(); void DataProcess(); matrix MatrixMul(matrix a, matrix b, int mod); matrix FastPow(matrix base, int n, int mod); int main() { while (~scanf("%d %d", &n, &m) && n && m) { InitRead(); DataProcess(); } return 0; } void InitRead() { for (int i=0; i<n; ++i) for (int j=0; j<m; ++j) scanf("%d", &a[i][j]); for (int i=0; i<m; ++i) for (int j=0; j<n; ++j) scanf("%d", &b[i][j]); return; } void DataProcess() { matrix tmp(m, m); for (int i=0; i<m; ++i) //计算B×A { for (int j=0; j<m; ++j) { tmp.mat[i][j] = 0; for (int k=0; k<n; ++k) { tmp.mat[i][j] += b[i][k] * a[k][j]; } tmp.mat[i][j] %= MOD; } } tmp = FastPow(tmp, n * n - 1, MOD); //计算(B×A)^(n×n-1) int temp[N][M]; for (int i=0; i<n; ++i) //计算A×(B×A)^(n×n-1) { for (int j=0; j<m; ++j) { temp[i][j] = 0; for (int k=0; k<m; ++k) { temp[i][j] += a[i][k] * tmp.mat[k][j]; } temp[i][j] %= MOD; } } int sum = 0; for (int i=0; i<n; ++i) //计算M { for (int j=0; j<n; ++j) { res[i][j] = 0; for (int k=0; k<m; ++k) { res[i][j] += temp[i][k] * b[k][j]; } res[i][j] %= MOD; sum += res[i][j]; } } printf("%d\n", sum); return; } matrix MatrixMul(matrix a, matrix b, int mod) { matrix ans(a.n, b.m); for (int i=0; i<a.n; ++i) { for (int j=0; j<b.m; ++j) { ans.mat[i][j] = 0; for (int k=0; k<a.m; ++k) { ans.mat[i][j] += a.mat[i][k] * b.mat[k][j]; } ans.mat[i][j] %= mod; } } return ans; } matrix FastPow(matrix base, int n, int mod) { matrix ans; ans.Unit(base.n); while (n) { if (n & 1) ans = MatrixMul(ans, base, mod); base = MatrixMul(base, base, mod); n >>= 1; } return ans; }