Find a specific number in an ordered array (binary search)

Question:
Given n elements arr[0:n-1] sorted, now we have to find a specific element x among these n elements. The
basic idea:
divide n elements into two halves of roughly the same number. Take arr[n/2] and compare with x. If x=arr[n/2], x is found and the algorithm terminates. If x<arr[n/2], just continue to search for x in the left half of the array arr. If x>arr[n/2], you only need to continue looking for x in the right half of the array arr.
Worst case time complexity: O(log n)

code show as below:

#define _CRT_SECURE_NO_WARNINGS 1
#include<stdio.h>
int Search(int x,int arr[],int n)
{
    
    
	int left = 0;//左下标
	int right = n-1;//右下标
	while (left <= right)
	{
    
    
		int mid = (left + right) / 2;
		if (x == arr[mid])
			return mid;
		if (x > arr[mid])
			left = mid + 1;
		else
			right = mid - 1;
	}
	return -1;//未找到x
}
int main()
{
    
    
	int num = 0;
	int arr[] = {
    
    0,1,2,3,4,5,6,7,8,9,10};
	printf("请输入你想查找的数字：   ");
	scanf("%d", &num);
	int sc = sizeof(arr)/sizeof(arr[0]);//求数组的长度（sizeof 计算字节的大小）
	int temp = Search(num,arr,sc);
	if (temp==-1)
		printf("已给数组中找不到x\n");
	else
		printf("已找到x  下标是：%d\n", temp);
	return 0;
}