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       <dc:date>2026-04-30T07:23:00+0800</dc:date>
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        <description>比赛

本周冯如杯，没有打比赛。

学习总结

容斥原理

容斥的一些理解：

我们能快速知道的是至少满足性质集合 $S$ 的个数 $f(S)$，而很多情况下 $f(S)$ 对相同的 $|S|$ 是相同的，这个时候计算贡献就需要乘上组合数，因为统计的是所有 $|S|$$f(S)$$|S|$$S$$$
\sum _{i=0}^n (-1)^i \binom ni f(i)
$$$S$$$
\sum _{i=1}^n (-1)^{i+1} \binom ni f(i)
$$$f(0)$$S$$|S| = 0$$S$$1$$0$$|S| = 0$$S$$0$$1$$E &gt; 3V-6$$K_5$$K_{3, 3}$</description>
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        <title>2020-2021:teams:i_dont_know_png:week_summary_1:potassium</title>
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        <description>比赛

无

学习总结

莫比乌斯反演

莫比乌斯反演：$g(n)=\sum_{d|n}f(d)$，则$f(n)=\mu *g$

$\epsilon(i)=[i=1]$在积性函数里扮演了类似于自然数中$1$的角色，为什么让$\epsilon$扮演自然数中$1$的角色呢，因为$(f*\epsilon)(n)=\sum_{d|n}f(\frac nd)\epsilon(d)=f(n)$。

$$id(i)=i$$

$$1(i)=1$$

$$\phi(i)=\text{多少个&lt;i且与i互质}$$

$$d(i)=i \text{约数个数}$$

$$\sigma(i)=i \text{约数个数和}$$

设$n=\sum_{i=1}^{m}p_i^{k_i}$，则

$$\mu(n)=\left\{\begin{aligned}&amp;1,&amp;n=1\\&amp;(-1)^m,&amp;\forall _ik_i=1\\&amp;0,&amp;\exists_ik_i\geq2 \end{aligned}\right.$$

狄利克雷卷积中，$1$的逆是$\mu$，即$1*\mu=\epsilon$$(1*\mu)(n)$$…</description>
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        <dc:date>2020-05-09T08:30:03+0800</dc:date>
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        <description>比赛

2020.05.01 Codeforces Round #638 (Div. 2)

2020.05.03 QkOI Round1

学习总结

无</description>
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