如果有人能指导我如何在这里找到最坏的情况，以及如何将 alg1 的输出 a 作为 n 的函数，我将不胜感激.谢谢！
If anyone could guide me to how you would find the worst-case here, and how to get the output a of alg1 as a function of n, I would be very grateful. Thanks!
推荐答案我们可以计算此代码执行的确切增量数.首先，让我们替换
We can compute the exact number of increments this code executes. First, let's replace
for k=t to o+t do与
for k=1 to o+1 do这个改动后，两个内循环是这样的
After this change, two inner loops looks like this
for t=1 to o do for k=1 to o+1 do这些循环的迭代次数显然是o*(o+1).总迭代次数可以通过以下方式计算:
The number of iterations of these loops is obviously o*(o+1). The overall number of iterations can be calculated in the following way:
当使用大 O 表示法时，我们可以排除多项式的系数和低阶项.因此，复杂度为O(n^3).
We can exclude coefficients and lower order terms of the polynomial when using big-O notation. Therefore, the complexity is O(n^3).