写一段用分治法求解矩阵乘法的java代码

好的，下面是用分治法求解矩阵乘法的Java代码：

public class MatrixMultiplication {
    public int[][] multiply(int[][] A, int[][] B) {
        int m = A.length, n = A[0].length, k = B[0].length;
        int[][] res = new int[m][k];
        matrixMultiply(A, B, res, 0, 0, 0, 0, m, n, k);
        return res;
    }

    private void matrixMultiply(int[][] A, int[][] B, int[][] C,
                                int rA, int cA, int rB, int cB,
                                int m, int n, int k) {
        if (n != k) {
            throw new IllegalArgumentException("Matrix size mismatch");
        }
        if (m == 1 && n == 1 && k == 1) {
            C[rA][cB] += A[rA][cA] * B[rB][cB];
            return;
        }
        int newM = m / 2, newN = n / 2, newK = k / 2;
        matrixMultiply(A, B, C, rA, cA, rB, cB, newM, newN, newK);
        matrixMultiply(A, B, C, rA, cA + newN, rB + newN, cB, newM, newN, newK);
        matrixMultiply(A, B, C, rA, cA, rB, cB + newK, newM, newN, newK);
        matrixMultiply(A, B, C, rA, cA + newN, rB + newN, cB + newK, newM, newN, newK);
        matrixMultiply(A, B, C, rA + newM, cA, rB, cB, newM, n - newN, newK);
        matrixMultiply(A, B, C, rA + newM, cA + newN, rB + newN, cB, newM, n - newN, newK);
        matrixMultiply(A, B, C, rA + newM, cA, rB, cB + newK, newM, n - newN, newK);
        matrixMultiply(A, B, C, rA + newM, cA + newN, rB + newN, cB + newK, newM, n - newN, newK);
    }
}

这个算法的时间复杂度是 $O(n^3)$，但是分治法的优化使得它可以有效地处理大规模矩阵乘法运算。