【每日一题】练习开始咯~~

1.从有序可重复数组中寻找一个元素首次出现的位置（由有序可知，可以使用二分查找来提高搜索速度，不过需要注意可重复性的特点，因此在这个基础上略微修改查找函数即可）

int find_num(int *data, int num, int low, int high){
	int mid;

	mid = (low+high)/2;
	if(high>low){
		if(*(data+mid)==num)
			return find_num(data, num, low, mid);
		else if(*(data+mid)>num)
			return find_num(data, num, low, mid-1);
		else
			return find_num(data, num, mid+1, high);
	}
	else{
		if(*(data+high)==num)
			return low;
		else
			return -1;
	}
}

int main(int argc, char *argv[]){
	int result, num=6;
	int testdata[]={1,1,1,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,6,6};

	result = find_num(testdata, num, 0, sizeof(testdata)/sizeof(int));
	result>=0?printf("find %d in testdata[%d]
", num, result):printf("not found
");

	system("pause");
	return 0;
}