写出下面问题的C代码:要求键盘输入数据。设a[0:n-1]是一个已排好序的数组。请改写二分搜索算法，使得当搜索元素x不在数组中时，返回小于x的最大元素的位置i和大于x的最小元素位置j。当搜索元素在数组中时，i和j相同，均为x在数组中的位置。并对自己的程序进行复杂性分析。

#include <stdio.h>
#include <limits.h>

// 函数声明
int binary_search(int arr[], int n, int x, int *less_than, int *greater_than);

int main() {
    int n, i;
    printf("Enter the size of the sorted array: ");
    scanf("%d", &n);
    
    // 输入排序好的整数数组
    int a[n];
    for (i = 0; i < n; ++i) {
        printf("Element %d: ", i);
        scanf("%d", &a[i]);
    }
    
    int target;
    printf("Enter the element to search: ");
    scanf("%d", &target);
    
    int less_than_index, greater_than_index;
    if (binary_search(a, n, target, &less_than_index, &greater_than_index)) {
        printf("Element not found, less than x at index %d and greater than x at index %d.\n",
               less_than_index, greater_than_index);
    } else {
        printf("Element found at position %d.\n", less_than_index);
    }

    return 0;
}

// 二分查找函数，同时更新小于和大于目标值的索引
int binary_search(int arr[], int n, int x, int *less_than, int *greater_than) {
    int low = 0, high = n - 1;

    while (low <= high) {
        int mid = low + (high - low) / 2;

        if (arr[mid] == x) {
            *less_than = mid;
            *greater_than = mid;
            return 1; // 成功找到并返回位置
        } else if (arr[mid] < x) {
            *less_than = mid + 1;
            low = mid + 1;
        } else {
            *greater_than = mid;
            high = mid - 1;
        }
    }

    return 0; // 如果没找到，返回0
}