hdu 4965 Fast Matrix Calculation(矩阵快速幂)

题目链接；hdu 4965 Fast Matrix Calculation

题目大意：给定两个矩阵A，B，分别为N*K和K*N；

1. 矩阵C = A*B
2. 矩阵M=CN∗N
3. 将矩阵M中的所有元素取模6，得到新矩阵M‘
4. 计算矩阵M’中所有元素的和

解题思路：因为矩阵C为N*N的矩阵，N最大为1000，就算用快速幂也超时，但是因为C = A*B, 所以CN∗N=ABAB…AB=AC′N∗N−1B，C‘ = B*A, 为K*K的矩阵，K最大为6，完全可以接受。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 1005;
const int MOD = 6;
typedef int Mat[maxn][maxn];

int N, K;
Mat A, B, X, Y, tmp;

void put (Mat x, int r, int c) {
    for (int i = 0; i < K; i++) {
        for (int j = 0;  j < K; j++)
            printf("%d ", x[i][j]);
        printf("\n");
    }
}

void mul_mat (Mat ret, Mat a, Mat b, int r, int t, int c) {
    memset(tmp, 0, sizeof(tmp));

    for (int k = 0; k < t; k++) {
        for (int i = 0; i < r; i++)
            for (int j = 0; j < c; j++)
                tmp[i][j] = (tmp[i][j] + a[i][k] * b[k][j]) % MOD;
    }
    memcpy(ret, tmp, sizeof(tmp));
}

void pow_mat (Mat ret, Mat x, int n) {
    memset(Y, 0, sizeof(Y));
    for (int i = 0; i < K; i++)
        Y[i][i] = 1;

    while (n) {
        if (n&1)
            mul_mat(Y, Y, x, K, K, K);
        mul_mat(x, x, x, K, K, K);
        n >>= 1;
    }
    memcpy(ret, Y, sizeof(Y));
}


void init () {
    for (int i = 0; i < N; i++)
        for (int j = 0; j < K; j++)
            scanf("%d", &A[i][j]);


    for (int i = 0; i < K; i++)
        for (int j = 0; j < N; j++)
            scanf("%d", &B[i][j]);
}

int main () {
    while (scanf("%d%d", &N, &K) == 2 && N + K) {
        init();

        mul_mat(X, B, A, K, N, K);


        pow_mat(X, X, N*N-1);

        mul_mat(X, A, X, N, K, K);
        mul_mat(X, X, B, N, K, N);

        int ans = 0;
        for (int i = 0; i < N; i++)
            for (int j = 0; j < N; j++)
                ans += X[i][j];
        printf("%d\n", ans);
    }
    return 0;
}