您已经正确计算出 a 值c> i 内循环的第一个迭代是：

You have correctly calculated that the value of a after the i-th iteration of the inner loop is:

其中 a_j0 是 a 在 j 次外循环。内循环的停止条件为：

Where a_j0 is the value of a at the start of the j-th outer loop. The stopping condition for the inner loop is:

可以将其解决为二次不等式：

Which can be solved as a quadratic inequality:

因此，内部循环大约为 O（sqrt（a_j0 / b））。 a 的 next 起始值满足：

Therefore the inner loop is approximately O(sqrt(a_j0 / b)). The next starting value of a satisfies:

大致缩放为 sqrt（2b * a_j0）。精确地计算时间复杂度将非常繁琐，因此让我们从此处开始应用上述近似值：

Scaling roughly as sqrt(2b * a_j0). It would be quite tedious to compute the time complexity exactly, so let's apply the above approximations from here on:

其中 a_n 替换 a_j0 和 t_n 是内部循环的运行时间–当然，总的时间复杂度只是 t_n 的总和。请注意，第一项由 n = 1 给出，并且 a 的输入值定义为 a_0 。

Where a_n replaces a_j0, and t_n is the run-time of the inner loop – and of course the total time complexity is just the sum of t_n. Note that the first term is given by n = 1, and that the input value of a is defined to be a_0.

在直接解决此重复项之前，请注意，由于第二项 t_2 已经与第一个 t_1 的平方根成比例，而后者在总和中占所有其他项。

Before directly solving this recurrence, note that since the second term t_2 is already proportional to the square root of the first t_1, the latter dominates all other terms in the sum.

因此，总时间复杂度仅为 O（sqrt（a / b）） 。

The total time complexity is therefore just O(sqrt(a / b)).

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更新：数值测试。

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Update: numerical tests.

请注意，由于 a 的所有值变化均与 b 成正比，因此所有循环条件均为也与 b 成正比，可以通过设置 b = 1 来规范化 并且只改变 a 。

Note that, since all changes in the value of a are proportional to b, and all loop conditions are also proportional to b, the function can be "normalized" by setting b = 1 and only varying a.

Javascript测试函数，它测量内部循环执行的次数：

Javascript test function, which measures the number of times that the inner loop executes:

function T(n)
{
   let t = 0, k = 0;
   while (n >= 1) {
      k = 1;
      while (n >= k) {
          n -= k;
          k++; t++;
      }
   }
   return t;
}

sqrt（n）$ c的图$ c>对 T（n）：

令人信服的直线，确认时间复杂度的确是一半。

A convincing straight line which confirms that the time complexity is indeed half-power.