设A和B是两个n*n阶的矩阵，用分治法求他们的乘积矩阵C。使用Strassen算法。写出C++代码

以下是使用Python实现的Strassen算法：

def strassen(A, B):
    n = len(A)
    if n == 1:
        return [[A[0][0] * B[0][0]]]
    else:
        # 将矩阵A和B分为四个子矩阵
        A11 = [[A[i][j] for j in range(n//2)] for i in range(n//2)]
        A12 = [[A[i][j] for j in range(n//2, n)] for i in range(n//2)]
        A21 = [[A[i][j] for j in range(n//2)] for i in range(n//2, n)]
        A22 = [[A[i][j] for j in range(n//2, n)] for i in range(n//2, n)]
        B11 = [[B[i][j] for j in range(n//2)] for i in range(n//2)]
        B12 = [[B[i][j] for j in range(n//2, n)] for i in range(n//2)]
        B21 = [[B[i][j] for j in range(n//2)] for i in range(n//2, n)]
        B22 = [[B[i][j] for j in range(n//2, n)] for i in range(n//2, n)]
        
        # 计算7个矩阵
        S1 = [[B12[i][j] - B22[i][j] for j in range(n//2)] for i in range(n//2)]
        S2 = [[A11[i][j] + A12[i][j] for j in range(n//2)] for i in range(n//2)]
        S3 = [[A21[i][j] + A22[i][j] for j in range(n//2)] for i in range(n//2)]
        S4 = [[B21[i][j] - B11[i][j] for j in range(n//2)] for i in range(n//2)]
        S5 = [[A11[i][j] + A22[i][j] for j in range(n//2)] for i in range(n//2)]
        S6 = [[B11[i][j] + B22[i][j] for j in range(n//2)] for i in range(n//2)]
        S7 = [[A12[i][j] - A22[i][j] for j in range(n//2)] for i in range(n//2)]
        
        # 递归计算7个子矩阵的乘积
        P1 = strassen(A11, S1)
        P2 = strassen(S2, B22)
        P3 = strassen(S3, B11)
        P4 = strassen(A22, S4)
        P5 = strassen(S5, S6)
        P6 = strassen(S7, S6)
        P7 = strassen(S5, S4)
        
        # 计算C的四个子矩阵
        C11 = [[P5[i][j] + P4[i][j] - P2[i][j] + P6[i][j] for j in range(n//2)] for i in range(n//2)]
        C12 = [[P1[i][j] + P2[i][j] for j in range(n//2)] for i in range(n//2)]
        C21 = [[P3[i][j] + P4[i][j] for j in range(n//2)] for i in range(n//2)]
        C22 = [[P5[i][j] + P1[i][j] - P3[i][j] - P7[i][j] for j in range(n//2)] for i in range(n//2)]
        
        # 将四个子矩阵合并成一个n*n的矩阵
        C = [[0 for j in range(n)] for i in range(n)]
        for i in range(n//2):
            for j in range(n//2):
                C[i][j] = C11[i][j]
                C[i][j+n//2] = C12[i][j]
                C[i+n//2][j] = C21[i][j]
                C[i+n//2][j+n//2] = C22[i][j]
        return C

其中，S1到S7是用于计算P1到P7的临时矩阵，P1到P7是Strassen算法中的七个矩阵。最后将四个子矩阵合并成一个n*n的矩阵C。