这是一个简单的O(1/n)算法。它甚至做了一些有趣的事情!

function foo(list input) {
  int m;
  double output;

  m = (1/ input.size) * max_value;  
  output = 0;
  for (int i = 0; i < m; i++)
    output+= random(0,1);

  return output;
}

O(1/n) is possible as it describes how the output of a function changes given increasing size of input. If we are using the function 1/n to describe the number of instructions a function executes then there is no requirement that the function take zero instructions for any input size. Rather, it is that for every input size, n above some threshold, the number of instructions required is bounded above by a positive constant multiplied by 1/n. As there is no actual number for which 1/n is 0, and the constant is positive, then there is no reason why the function would constrained to take 0 or fewer instructions.

inline void O0Algorithm() {}

我经常用O(1/n)来描述随着输入变大而变小的概率——例如，在log2(n)次投掷中，一枚均匀硬币背面朝上的概率是O(1/n)。

这是一个简单的O(1/n)算法。它甚至做了一些有趣的事情!

function foo(list input) {
  int m;
  double output;

  m = (1/ input.size) * max_value;  
  output = 0;
  for (int i = 0; i < m; i++)
    output+= random(0,1);

  return output;
}

O(1/n) is possible as it describes how the output of a function changes given increasing size of input. If we are using the function 1/n to describe the number of instructions a function executes then there is no requirement that the function take zero instructions for any input size. Rather, it is that for every input size, n above some threshold, the number of instructions required is bounded above by a positive constant multiplied by 1/n. As there is no actual number for which 1/n is 0, and the constant is positive, then there is no reason why the function would constrained to take 0 or fewer instructions.

我相信量子算法可以通过叠加“一次”进行多次计算……

我怀疑这是一个有用的答案。

你不能低于O(1)但是O(k) k小于N是可能的。我们称之为次线性时间算法。在某些问题中，次线性时间算法只能给出特定问题的近似解。然而，有时，一个近似解就可以了，可能是因为数据集太大了，或者计算所有数据的计算成本太高了。