C语言编程题计算两个矩阵的和

1、使用双重循环逐元素相加

最基本的方式，逐元素相加两个矩阵。

#include <stdio.h>

void addMatrices(int rows, int cols, int mat1[rows][cols], int mat2[rows][cols], int result[rows][cols]) {
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < cols; j++) {
            result[i][j] = mat1[i][j] + mat2[i][j];
        }
    }
}

int main() {
    int rows = 2, cols = 2;
    int mat1[2][2] = {{1, 2}, {3, 4}};
    int mat2[2][2] = {{5, 6}, {7, 8}};
    int result[2][2];

    addMatrices(rows, cols, mat1, mat2, result);

    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < cols; j++) {
            printf("%d ", result[i][j]);
        }
        printf("\n");
    }

    return 0;
}

2、通过指针相加

使用指针来遍历矩阵并计算其和。

#include <stdio.h>

void addMatrices(int rows, int cols, int *mat1, int *mat2, int *result) {
    for (int i = 0; i < rows * cols; i++) {
        *(result + i) = *(mat1 + i) + *(mat2 + i);
    }
}

int main() {
    int rows = 2, cols = 2;
    int mat1[2][2] = {{1, 2}, {3, 4}};
    int mat2[2][2] = {{5, 6}, {7, 8}};
    int result[2][2];

    addMatrices(rows, cols, (int *)mat1, (int *)mat2, (int *)result);

    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < cols; j++) {
            printf("%d ", result[i][j]);
        }
        printf("\n");
    }

    return 0;
}

3、使用递归方式

通过递归计算矩阵和。

#include <stdio.h>

void addMatricesRecursive(int rows, int cols, int mat1[rows][cols], int mat2[rows][cols], int result[rows][cols], int i, int j) {
    if (i == rows) {
        return;
    }
    if (j == cols) {
        addMatricesRecursive(rows, cols, mat1, mat2, result, i + 1, 0);
        return;
    }
    result[i][j] = mat1[i][j] + mat2[i][j];
    addMatricesRecursive(rows, cols, mat1, mat2, result, i, j + 1);
}

int main() {
    int rows = 2, cols = 2;
    int mat1[2][2] = {{1, 2}, {3, 4}};
    int mat2[2][2] = {{5, 6}, {7, 8}};
    int result[2][2];

    addMatricesRecursive(rows, cols, mat1, mat2, result, 0, 0);

    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < cols; j++) {
            printf("%d ", result[i][j]);
        }
        printf("\n");
    }

    return 0;
}