题目如下：

解题思路：几乎和 的方法一样。只不过一个是减法一个是乘法，还有一点区别是中i-j和j-i只算一个元素，而本题中i*j与j*i算两个元素。

代码如下：

class Solution(object):    def findKthNumber(self, m, n, k):        """        :type m: int        :type n: int        :type k: int        :rtype: int        """        low,high = 1, m*n        while low <= high:            mid = (low + high) / 2            less,equal = 0,0            for i in range(1,min(mid+1,m+1)):                less += min(mid/i,n)                if mid/i <= n and mid % i == 0:                    equal += 1                    less -= 1            if less >= k:                high = mid - 1            elif less + equal < k:                low = mid + 1            elif less == k and equal == 0:                high = mid - 1            else:                break        return mid