内部循环将准确执行0 +1 + ... + n ^ 2-2 + n ^ 2-1-=(n ^ 2)(n ^ 2-1)/2次(请参见算术系列)，因此实际上是O(n ^ 4).

The inner loop will execute exactly 0 + 1 + ... + n^2 - 2 + n^2 - 1 = (n^2)(n^2 - 1)/2 times (see Arithmetic Series), so it's actually O(n^4).