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：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int get_first_k(int *, int, int, int);
int get_last_k(int *, int, int, int);

int main(int argc, char *argv[])
{
    int a[13] = {1, 2, 5, 5, 5, 5, 16, 20, 23, 24, 28, 30, 43};
    int b = 430; 
    int num = 0;
    int first = 0;
    int last = 0;

    first = get_first_k(a, b, 0, sizeof(a)/sizeof(int) - 1);
    if (first == -1)
    {
        printf("不存在该数字
");
        return 0;
    }

    last = get_last_k(a, b, 0, sizeof(a)/sizeof(int) - 1);
    printf("%d  %d
", first, last);    
    num = last - first + 1;
    printf("%d
", num);


    return 0;
}
/*
int search(int *a, int target, int p, int r)
{
    if (p <= r)
    {
        int mid;

        mid = (p + r) / 2; 
        if (*(a + mid) == target)
            return 1;
        else if (*(a + mid) > target)
            return search(a, target, p, mid - 1);
        else
            return search(a, target, mid + 1, r);
    }
    return 0;
}
*/
int get_first_k(int *a, int target, int p, int r)
{
    int mid = 0;

    while (p <= r)
    {
        mid = (p + r) / 2;
        if (*(a + mid) == target)
        {
            if (mid == 0 || *(a + mid - 1) != target)
            {
                return mid;
            }
            else
            {
                r = mid - 1;
            }
        }
        else if (*(a + mid) > target)
        {
            r = mid - 1;
        }
        else
        {
            p = mid + 1;
        }

    }

    return -1;
}

int get_last_k(int *a, int target, int p, int r)
{
    int mid = 0;
    int len = r - p + 1;

    while (p <= r)
    {
        mid = (p + r) / 2;
        if (*(a + mid) == target)
        {
            if (mid == len - 1 || *(a + mid + 1) != target)
            {
                return mid;
            }
            else
            {
                p = mid + 1;
            }
        }
        else if (*(a + mid) > target)
        {
            r = mid - 1;
        }
        else
        {
            p = mid + 1;
        }
    }
    return -1;
}