题目描述：
You have a total of n coins that you want to form in a staircase shape, where every k-th row must have exactly k coins.
Given n, find the total number of full staircase rows that can be formed.
n is a non-negative integer and fits within the range of a 32-bit signed integer.
Example 1:
n = 5 The coins can form the following rows: ¤ ¤ ¤ ¤ ¤ Because the 3rd row is incomplete, we return 2.
Example 2:
n = 8
The coins can form the following rows:
¤
¤ ¤
¤ ¤ ¤
¤ ¤
Because the 4th row is incomplete, we return 3.
要完成的函数：
int arrangeCoins(int n)
说明：
1、这道题目看起来很容易，换成人类来思考的话，就是利用等差数列的和来做，满足t(t+1)<=2n<(t+1)(t+2)，然后返回t。
我们对上面式子做个变形，要由n求出t值，很容易的。
t(t+1)<=2n
t^2+t<=2n
t^2+t+1/4<=2n+1/4
(t+1/2)^2<=2n+1/4
t<=sqrt(2n+1/4)-1/2
t是最靠近的那个整数，所以取下地板函数floor()
代码如下：
int arrangeCoins(int n)
{
return floor(sqrt((double)2*n+0.25)-0.5);
}
上述代码实测41ms，排名很靠后。
2、改进：
上述代码浪费时间的地方在于sqrt函数，sqrt内部实现应该是泰勒展开，比较耗时间。
如果我们把泰勒公式换成二分查找，应该会更快，几趟迭代就能够找到我们要的值。
改进完实测33ms，beats 78.81% of cpp submissions。
二分查找很熟悉啦，而且前面做过差不多的题目，就不再贴出来了。
具体可以参考leetcode网站：https://leetcode.com/problems/arranging-coins/discuss/92314/C++-Three-solutions:-O(n)-O(logn)-O(1)