用C语言写出使用分治法解决规模为n*n(在主函数中手动输入大于2小于100）矩阵乘法的代码

好的，以下是使用分治法解决规模为n*n的矩阵乘法的C语言代码：

#include <stdio.h>

void matrix_multiply(int n, int A[][n], int B[][n], int C[][n]) {
    if (n == 1) {
        C[0][0] = A[0][0] * B[0][0];
    } else {
        int i, j;
        int half_size = n / 2;
        int A11[half_size][half_size], A12[half_size][half_size], A21[half_size][half_size], A22[half_size][half_size];
        int B11[half_size][half_size], B12[half_size][half_size], B21[half_size][half_size], B22[half_size][half_size];
        int C11[half_size][half_size], C12[half_size][half_size], C21[half_size][half_size], C22[half_size][half_size];
        int P1[half_size][half_size], P2[half_size][half_size], P3[half_size][half_size], P4[half_size][half_size], P5[half_size][half_size], P6[half_size][half_size], P7[half_size][half_size];
        int temp1[half_size][half_size], temp2[half_size][half_size];

        for (i = 0; i < half_size; i++) {
            for (j = 0; j < half_size; j++) {
                A11[i][j] = A[i][j];
                B11[i][j] = B[i][j];
            }
        }
        for (i = 0; i < half_size; i++) {
            for (j = half_size; j < n; j++) {
                A12[i][j-half_size] = A[i][j];
                B12[i][j-half_size] = B[i][j];
            }
        }
        for (i = half_size; i < n; i++) {
            for (j = 0; j < half_size; j++) {
                A21[i-half_size][j] = A[i][j];
                B21[i-half_size][j] = B[i][j];
            }
        }
        for (i = half_size; i < n; i++) {
            for (j = half_size; j < n; j++) {
                A22[i-half_size][j-half_size] = A[i][j];
                B22[i-half_size][j-half_size] = B[i][j];
            }
        }

        // Calculate seven products
        matrix_multiply(half_size, A11, B11, P1);
        matrix_multiply(half_size, A12, B21, P2);
        matrix_multiply(half_size, A11, B12, P3);
        matrix_multiply(half_size, A12, B22, P4);
        matrix_multiply(half_size, A21, B11, P5);
        matrix_multiply(half_size, A22, B21, P6);
        matrix_multiply(half_size, A21, B12, P7);
        matrix_multiply(half_size, A22, B22, temp1);

        // Combine products to form C matrix
        for (i = 0; i < half_size; i++) {
            for (j = 0; j < half_size; j++) {
                C11[i][j] = P1[i][j] + P2[i][j];
                C12[i][j] = P3[i][j] + P4[i][j];
                C21[i][j] = P5[i][j] + P6[i][j];
                C22[i][j] = P7[i][j] + temp1[i][j];
            }
        }

        // Copy C11, C12, C21, and C22 into C matrix
        for (i = 0; i < half_size; i++) {
            for (j = 0; j < half_size; j++) {
                C[i][j] = C11[i][j];
                C[i][j+half_size] = C12[i][j];
                C[i+half_size][j] = C21[i][j];
                C[i+half_size][j+half_size] = C22[i][j];
            }
        }
    }
}

int main() {
    int n, i, j;
    printf("Please enter the size of the matrix (n*n, n>2,n<100): ");
    scanf("%d", &n);
    int A[n][n], B[n][n], C[n][n];

    printf("Please enter elements of matrix A: ");
    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
            scanf("%d", &A[i][j]);
        }
    }

    printf("Please enter elements of matrix B: ");
    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
            scanf("%d", &B[i][j]);
        }
    }

    matrix_multiply(n, A, B, C);

    printf("Result matrix: \n");
    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
            printf("%d ", C[i][j]);
        }
        printf("\n");
    }

    return 0;
}