Fast Matrix Calculation HDU

原题链接

• 题意：给出矩阵 (A)，为 (N imes M) 的矩阵，矩阵 (B) 为 (M imes N) 的矩阵，(4 <= N <= 1000, 1 <= M <= 10) ，设矩阵 (C = A imes B) 求出 (C^n) 各个元素和。
• 题解：可以发现的是乘一步 (O(n^3)) 显然吃不消。但是，可以发现一个性质，即 (M) 比较小。((A imes B) imes(A imes B) = A imes(B imes A) imes B) 可以发现，如果 (n) 个 (A imes B) 相乘，可以转化成 (n-1) 个 ((B imes A)) 相乘，设结果为矩阵 (T)，那么最终结果即为 (A imes T imes B), 而求 (B imes A) 是 (N imes M) 和 (M imes N) 是 (O(m^2n)) 的复杂度，可以接受。算 (T) 的复杂度是可用快速幂，即 (O(m^2n imes log n^2)),然后最终 (A imes T) 是 (O(m^2n)) 的复杂度，然后再乘 (B) 是 (O(n^2m)) 的复杂度。
  最终复杂的度是 (O(m^2n imes log n^2 + n^2m))。
• 代码：

#include <iostream>
#include <cstring>

using namespace std;


typedef long long ll;
const int N = 1e3  + 9;
const ll mod = 6;
int n, m;
ll A[N][N], B[N][N], T[N][N],ans[N][N];
struct Matrix {
    ll a[10][10];
    Matrix(){memset(a, 0, sizeof a);}
    Matrix operator*(Matrix rhs)const {
        Matrix ret;
        for (int i = 1; i <= m; i ++) {
            for (int j = 1; j <= m; j++){
                for (int k = 1; k <= m; k++) {
                    (ret.a[i][j] += (a[i][k] * rhs.a[k][j] % mod)) %= mod;
                }
            }
        }
        return ret;
    }
    void pr() {
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= m; j++) {
                cout << a[i][j] << " ";
            }
            cout << endl;
        }
    }
};
Matrix ksm (Matrix A, int kk) {
    if (kk == 1)return A;
    Matrix ret;
    bool f = 0;
    //cout << kk << "???";
    while (kk) {
        if (kk & 1) {
            if (!f) {
                ret = A;
                f = 1;
               // cout << "?";
            } else
            ret = ret * A;
        }
        kk >>= 1;
        A = A * A;
    }
    return ret;
}
void solve() {
while (cin >> n >> m) {

    if (n == 0 && m == 0)return;
    Matrix C;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> A[i][j];
        }
    } 
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            cin >> B[i][j];
        }
    }
    for (int i = 1; i <= m; i ++) {
        for (int j = 1; j <= m; j ++) {
            for (int k = 1; k <= n; k++) {
                (C.a[i][j] += B[i][k] * A[k][j] % mod)%=mod;
            }
        }
    }
    int kk = n * n-1;
    Matrix M = ksm(C, kk);
    memset(T, 0, sizeof T);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            for (int k = 1; k <= m; k++) {
                (T[i][j] += A[i][k] * M.a[k][j] % mod) %= mod;
            }
        }
    }
    ll sum = 0;
    memset(ans, 0, sizeof ans);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            for (int k = 1; k <= m; k++) {
                (ans[i][j] += T[i][k] * B[k][j] % mod) %= mod;
            }
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            sum += ans[i][j];
        }
    }   
    cout << sum << endl;
}

}
signed main() {
    int t = 1;//cin >> t;
    while (t--) {
        solve();
    }
}