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二分查找法：从一个有序数组中找到一个数（不管是升序还是降序）。
思路：先从中间找，之后依次截半查找。
比如从1 2 3 4 5中找到4，先找中间数3，发现4小那么取3与5最中间的数4，找到了。
如果是1 2 3 4找3，先找下标为(0 + 3)/2，就是下标为1的数2，发现比3小，再取3与4中间下标(2 + 3)/2，就是下标
为2的数3，找到了（这里的下标是访问数组元素的下标）。
（尽力了，给自己点亿点痛苦）
代码如下：
#include<stdio.h> int main(void) { int arr[10] = { 1,2,3,4,5,6,7,8,9,10 }; int
k = 0; scanf("%d", &k);//输入查找数 int left = 0; int right = sizeof(arr) /
sizeof(arr[0]) - 1;//下标不能为10，最大为9 while (left <= right)//最坏打算 = 找了半天竟是两端的数 {
int mid = (left + right) / 2; if (k > arr[mid]) { left = mid + 1; } else if (k
< arr[mid]) { right = mid - 1; } else { printf("找到了\n"); break; } } if (left >
right) { printf("没找到\n"); } return 0; }
如有错误可直接指出，希望编程小伙伴们能一起进步成长。
