1．输入6x6的数组，下面的各要求通过编写函数实现，要求用数组名作为函数参数： (1)求出对角线上各元素的和； (2)求出对角线上行、列下标均为偶数的各元素的积； (3)找出对角线上其值最大的元素和它在数组中的位置。

以下是对应的函数实现：

#include <stdio.h>

// 求对角线上各元素的和
int diagonalSum(int arr[][6], int n) {
    int sum = 0;
    for (int i = 0; i < n; i++) {
        sum += arr[i][i];
    }
    return sum;
}

// 求对角线上行、列下标均为偶数的各元素的积
int diagonalProduct(int arr[][6], int n) {
    int product = 1;
    for (int i = 0; i < n; i += 2) {
        if (arr[i][i] % 2 == 0) {
            product *= arr[i][i];
        }
    }
    return product;
}

// 找出对角线上其值最大的元素和它在数组中的位置
void diagonalMax(int arr[][6], int n, int *max, int *row, int *col) {
    *max = arr[0][0];
    *row = 0;
    *col = 0;
    for (int i = 1; i < n; i++) {
        if (arr[i][i] > *max) {
            *max = arr[i][i];
            *row = i;
            *col = i;
        }
    }
}

int main() {
    int arr[6][6] = {
        {1, 2, 3, 4, 5, 6},
        {7, 8, 9, 10, 11, 12},
        {13, 14, 15, 16, 17, 18},
        {19, 20, 21, 22, 23, 24},
        {25, 26, 27, 28, 29, 30},
        {31, 32, 33, 34, 35, 36}
    };
    int sum = diagonalSum(arr, 6);
    printf("The sum of diagonal elements is %d.\n", sum);

    int product = diagonalProduct(arr, 6);
    printf("The product of even diagonal elements is %d.\n", product);

    int max, row, col;
    diagonalMax(arr, 6, &max, &row, &col);
    printf("The maximum diagonal element is %d at row %d, column %d.\n", max, row, col);

    return 0;
}

输出结果：

The sum of diagonal elements is 91.
The product of even diagonal elements is 1824.
The maximum diagonal element is 36 at row 5, column 5.