分治法实现矩阵的乘法，用c++来实现

以下是使用分治法实现矩阵乘法的C代码：

#include <stdio.h>

#define N 4

// 分治法实现矩阵乘法
void matrix_multiply(int a[][N], int b[][N], int c[][N], int n) {
    if (n == 1) {
        c[0][0] = a[0][0] * b[0][0];
    } else {
        int a11[N/2][N/2], a12[N/2][N/2], a21[N/2][N/2], a22[N/2][N/2];
        int b11[N/2][N/2], b12[N/2][N/2], b21[N/2][N/2], b22[N/2][N/2];
        int c11[N/2][N/2], c12[N/2][N/2], c21[N/2][N/2], c22[N/2][N/2];
        int i, j;

        // 划分矩阵
        for (i = 0; i < n/2; i++) {
            for (j = 0; j < n/2; j++) {
                a11[i][j] = a[i][j];
                a12[i][j] = a[i][j+n/2];
                a21[i][j] = a[i+n/2][j];
                a22[i][j] = a[i+n/2][j+n/2];

                b11[i][j] = b[i][j];
                b12[i][j] = b[i][j+n/2];
                b21[i][j] = b[i+n/2][j];
                b22[i][j] = b[i+n/2][j+n/2];
            }
        }

        // 递归计算乘积矩阵
        matrix_multiply(a11, b11, c11, n/2);
        matrix_multiply(a12, b21, c12, n/2);
        matrix_multiply(a11, b12, c21, n/2);
        matrix_multiply(a12, b22, c22, n/2);

        // 合并结果
        for (i = 0; i < n/2; i++) {
            for (j = 0; j < n/2; j++) {
                c[i][j] = c11[i][j] + c12[i][j];
                c[i][j+n/2] = c21[i][j] + c22[i][j];
                c[i+n/2][j] = c11[i][j+n/2] + c12[i+n/2][j+n/2];
                c[i+n/2][j+n/2] = c21[i+n/2][j] + c22[i+n/2][j+n/2];
            }
        }
    }
}

int main() {
    int a[N][N] = {
        {1, 2, 3, 4},
        {5, 6, 7, 8},
        {9, 10, 11, 12},
        {13, 14, 15, 16}
    };
    int b[N][N] = {
        {1, 0, 0, 0},
        {0, 1, 0, 0},
        {0, 0, 1, 0},
        {0, 0, 0, 1}
    };
    int c[N][N];
    int i, j;

    matrix_multiply(a, b, c, N);

    // 输出结果
    for (i = 0; i < N; i++) {
        for (j = 0; j < N; j++) {
            printf("%d ", c[i][j]);
        }
        printf("\n");
    }

    return 0;
}

该程序首先定义了一个4x4的矩阵a和一个4x4的单位矩阵b。然后调用matrix_multiply函数来计算它们的乘积。最后输出结果。

matrix_multiply函数使用了递归的方式来实现矩阵乘法。当矩阵大小为1时，直接计算乘积。否则，将矩阵分成4个大小为n/2的子矩阵，并递归计算它们的乘积，最后合并结果。

注意，该程序中只实现了矩阵大小为2的幂次方的乘法，如果要计算其他大小的矩阵乘法，需要对代码进行修改。