数字在排序数组中出现的次数

题目：统计一个数字在排序数组中出现的次数．例如，输入排序数组｛1，2，3，3，3，3，4，5｝和数字３，由于在这个数组中出现了４次，因此输出４．

使用二分查找，基本思想：先查找该数字第一次出现的位置，然后查找该数字最后一次出现的位置．代码如下：

  1 #include <stdio.h>
  2 #include <stdlib.h>
  3 #include <string.h>
  4 
  5 int get_first_k(int *, int, int, int);
  6 int get_last_k(int *, int, int, int);
  7 
  8 int main(int argc, char *argv[])
  9 {
 10     int a[13] = {1, 2, 5, 5, 5, 5, 16, 20, 23, 24, 28, 30, 43};
 11     int b = 430; 
 12     int num = 0;
 13     int first = 0;
 14     int last = 0;
 15 
 16     first = get_first_k(a, b, 0, sizeof(a)/sizeof(int) - 1);
 17     if (first == -1)
 18     {
 19         printf("不存在该数字
");
 20         return 0;
 21     }
 22 
 23     last = get_last_k(a, b, 0, sizeof(a)/sizeof(int) - 1);
 24     printf("%d  %d
", first, last);    
 25     num = last - first + 1;
 26     printf("%d
", num);
 27 
 28 
 29     return 0;
 30 }
 31 /*
 32 int search(int *a, int target, int p, int r)
 33 {
 34     if (p <= r)
 35     {
 36         int mid;
 37 
 38         mid = (p + r) / 2; 
 39         if (*(a + mid) == target)
 40             return 1;
 41         else if (*(a + mid) > target)
 42             return search(a, target, p, mid - 1);
 43         else
 44             return search(a, target, mid + 1, r);
 45     }
 46     return 0;
 47 }
 48 */
 49 int get_first_k(int *a, int target, int p, int r)
 50 {
 51     int mid = 0;
 52 
 53     while (p <= r)
 54     {
 55         mid = (p + r) / 2;
 56         if (*(a + mid) == target)
 57         {
 58             if (mid == 0 || *(a + mid - 1) != target)
 59             {
 60                 return mid;
 61             }
 62             else
 63             {
 64                 r = mid - 1;
 65             }
 66         }
 67         else if (*(a + mid) > target)
 68         {
 69             r = mid - 1;
 70         }
 71         else
 72         {
 73             p = mid + 1;
 74         }
 75 
 76     }
 77 
 78     return -1;
 79 }
 80 
 81 int get_last_k(int *a, int target, int p, int r)
 82 {
 83     int mid = 0;
 84     int len = r - p + 1;
 85 
 86     while (p <= r)
 87     {
 88         mid = (p + r) / 2;
 89         if (*(a + mid) == target)
 90         {
 91             if (mid == len - 1 || *(a + mid + 1) != target)
 92             {
 93                 return mid;
 94             }
 95             else
 96             {
 97                 p = mid + 1;
 98             }
 99         }
100         else if (*(a + mid) > target)
101         {
102             r = mid - 1;
103         }
104         else
105         {
106             p = mid + 1;
107         }
108     }
109     return -1;
110 }