调和级数的复杂度

for (int i = 1;i <= n;i++)
	for (int j = 1;j <= n/i;j++) ...

是调和级数级别的复杂度，可以由调和级数近似求和公式得到

(T(n)=nsum_{i=1}^n frac{1}{n}=O(n ln n))

具体过程如下

[egin{eqnarray*} && sum_{i=1}^nfrac{1}{i} \ && =sum_{i=1}^nint_i^{i+1}frac{1}{lfloor x floor}dx \ && =int_1^{n+1}frac{1}{x}-frac{1}{x}+frac{1}{lfloor x floor}dx \ && =int_1^{n+1}frac{1}{x}dx+int_1^{n+1}frac{1}{lfloor x floor}-frac{1}{x}dx \ && approx ln(n+1)+gamma end{eqnarray*}]

其中 (gamma approx 0.5772156649)