注：所以时间均在[Judge Duck Online](https://duck.ac/problem/mmmd1k)测试，用一维数组存储，其实二维数组储存也是一样的，因为地址都是按行连续的，按测试时间排序。 ```cpp //ikj:707.201 ms #define idx(i, j) ((i) * n + (j)) void matrix_multiply(int n, const double *A, const double *B, double *C) { for (int i = 0; i < n; i++) for (int k = 0; k < n; k++) for (int j = 0; j < n; j++) C[idx(i, j)] += A[idx(i, k)] * B[idx(k, j)]; } ``` ```cpp //kij:860.137 ms #define idx(i, j) ((i) * n + (j)) void matrix_multiply(int n, const double *A, const double *B, double *C) { for (int k = 0; k < n; k++) for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) C[idx(i, j)] += A[idx(i, k)] * B[idx(k, j)]; } ``` ```cpp //ijk:6.212 s #define idx(i, j) ((i) * n + (j)) void matrix_multiply(int n, const double *A, const double *B, double *C) { for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) for (int k = 0; k < n; k++) C[idx(i, j)] += A[idx(i, k)] * B[idx(k, j)]; } ``` ```cpp //jik:8.07 s #define idx(i, j) ((i) * n + (j)) void matrix_multiply(int n, const double *A, const double *B, double *C) { for (int j = 0; j < n; j++) for (int i = 0; i < n; i++) for (int k = 0; k < n; k++) C[idx(i, j)] += A[idx(i, k)] * B[idx(k, j)]; } ``` ```cpp //jki:>10 s #define idx(i, j) ((i) * n + (j)) void matrix_multiply(int n, const double *A, const double *B, double *C) { for (int j = 0; j < n; j++) for (int k = 0; k < n; k++) for (int i = 0; i < n; i++) C[idx(i, j)] += A[idx(i, k)] * B[idx(k, j)]; } ``` ```cpp //kji:>10 s #define idx(i, j) ((i) * n + (j)) void matrix_multiply(int n, const double *A, const double *B, double *C) { for (int k = 0; k < n; k++) for (int j = 0; j < n; j++) for (int i = 0; i < n; i++) C[idx(i, j)] += A[idx(i, k)] * B[idx(k, j)]; } ``` 发现第三重循环如果需要用的数在同一行中速度可以达到最快，ikj 和 kij 都是较快的方法 ## 参考 - [C++加速矩阵乘法的最简单方法 - 知乎](https://zhuanlan.zhihu.com/p/146250334) Loading... 注：所以时间均在[Judge Duck Online](https://duck.ac/problem/mmmd1k)测试，用一维数组存储，其实二维数组储存也是一样的，因为地址都是按行连续的，按测试时间排序。 ```cpp //ikj:707.201 ms #define idx(i, j) ((i) * n + (j)) void matrix_multiply(int n, const double *A, const double *B, double *C) { for (int i = 0; i < n; i++) for (int k = 0; k < n; k++) for (int j = 0; j < n; j++) C[idx(i, j)] += A[idx(i, k)] * B[idx(k, j)]; } ``` ```cpp //kij:860.137 ms #define idx(i, j) ((i) * n + (j)) void matrix_multiply(int n, const double *A, const double *B, double *C) { for (int k = 0; k < n; k++) for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) C[idx(i, j)] += A[idx(i, k)] * B[idx(k, j)]; } ``` ```cpp //ijk:6.212 s #define idx(i, j) ((i) * n + (j)) void matrix_multiply(int n, const double *A, const double *B, double *C) { for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) for (int k = 0; k < n; k++) C[idx(i, j)] += A[idx(i, k)] * B[idx(k, j)]; } ``` ```cpp //jik:8.07 s #define idx(i, j) ((i) * n + (j)) void matrix_multiply(int n, const double *A, const double *B, double *C) { for (int j = 0; j < n; j++) for (int i = 0; i < n; i++) for (int k = 0; k < n; k++) C[idx(i, j)] += A[idx(i, k)] * B[idx(k, j)]; } ``` ```cpp //jki:>10 s #define idx(i, j) ((i) * n + (j)) void matrix_multiply(int n, const double *A, const double *B, double *C) { for (int j = 0; j < n; j++) for (int k = 0; k < n; k++) for (int i = 0; i < n; i++) C[idx(i, j)] += A[idx(i, k)] * B[idx(k, j)]; } ``` ```cpp //kji:>10 s #define idx(i, j) ((i) * n + (j)) void matrix_multiply(int n, const double *A, const double *B, double *C) { for (int k = 0; k < n; k++) for (int j = 0; j < n; j++) for (int i = 0; i < n; i++) C[idx(i, j)] += A[idx(i, k)] * B[idx(k, j)]; } ``` 发现第三重循环如果需要用的数在同一行中速度可以达到最快，ikj 和 kij 都是较快的方法 ## 参考 - [C++加速矩阵乘法的最简单方法 - 知乎](https://zhuanlan.zhihu.com/p/146250334) 最后修改：2025 年 07 月 02 日 © 允许规范转载 赞 1 如果觉得我的文章对你有用，请随意赞赏