莫比乌斯反演总结

1|0[国家集训队] Crash的数字表格 / JZPTAB

$$\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}\operatorname{lcm}(i, j)\\ = \sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}\sum\limits_{k |i, j}[\gcd(i, j) = k] \times \frac{ij}{k}\\ = \sum\limits_{k}^{\min(n, m)}k\times\sum\limits_{i = 1}^{\frac{n}{k}}i\sum\limits_{j = 1}^{\frac{m}{k}}j\times[gcd(i, j) = 1]\\ = \sum\limits_{k}^{\min(n, m)}k\times\sum\limits_{i = 1}^{\frac{n}{k}}i\sum\limits_{j = 1}^{\frac{m}{k}}j\sum\limits_{e | i, j}\mu(e)\\ = \sum\limits_{k}^{\min(n, m)}k\times\sum\limits_{e}^{\min(\frac{n}{k}, \frac{m}{k})}\mu(e)\times\sum\limits_{i = 1}^{\frac{n}{ke}}\sum\limits_{j = 1}^{\frac{m}{ke}}ij\\ = \sum\limits_{k}^{\min(n, m)}k\times\sum\limits_{e}^{\min(\frac{n}{k}, \frac{m}{k})}\mu(e)\frac{(1 + \lfloor\frac{n}{ke}\rfloor) \times \lfloor\frac{n}{ke}\rfloor}{2}\frac{(1 + \lfloor\frac{m}{ke}\rfloor) \times \lfloor\frac{m}{ke}\rfloor}{2}\\ $$

数论分块套数论分块即可

2|0约数个数和

$$\sum\limits_{i=1}^n\sum\limits_{j=1}^md(ij)\\ = \sum\limits_{i=1}^n\sum\limits_{j=1}^m\sum\limits_{x|i}\sum\limits_{y|j}[\gcd(x, y) = 1]\\ = \sum\limits_{x}^{n}\sum\limits_{y}^{m}[\gcd(x, y) = 1]\sum\limits_{i}^{\frac{n}{x}}\sum\limits_{j}^{\frac{m}{y}}1\\ = \sum\limits_{x}^{n}\sum\limits_{y}^{m}[\gcd(x, y) = 1]\lfloor\frac{n}{x}\rfloor\lfloor\frac{n}{y}\rfloor\\ = \sum\limits_{x}^{n}\sum\limits_{y}^{m}\lfloor\frac{n}{x}\rfloor\lfloor\frac{n}{y}\rfloor[\gcd(x, y) = 1]\\ = \sum\limits_{x}^{n}\sum\limits_{y}^{m}\lfloor\frac{n}{x}\rfloor\lfloor\frac{m}{y}\rfloor\sum_{i|x, y}\mu(i)\\ = \sum_{i}^{\min(n, m)}\mu(i)\sum\limits_{x}^{\frac{n}{i}}\sum\limits_{y}^{\frac{m}{i}}\lfloor\frac{n}{xi}\rfloor\lfloor\frac{m}{yi}\rfloor\\ = \sum_{i}^{\min(n, m)}\mu(i)\sum\limits_{x}^{\frac{n}{i}}\lfloor\frac{n}{xi}\rfloor\sum\limits_{y}^{\frac{m}{i}}\lfloor\frac{m}{yi}\rfloor\\ = \sum_{i}^{\min(n, m)}\mu(i)(\sum\limits_{x}^{\frac{n}{i}}\lfloor\frac{n}{xi}\rfloor)(\sum\limits_{y}^{\frac{m}{i}}\lfloor\frac{m}{yi}\rfloor)\\ $$

预处理 $\sum\limits_{j = 1}^{i}\lfloor\frac{i}{j}\rfloor$ 再套数论分块即可

3|0ZAP-Queries

$$\sum\limits_{i = 1}^{a}\sum\limits_{j = 1}^{b}[\gcd(i, j) = d]\\ = \sum\limits_{i = 1}^{\frac{a}{d}}\sum\limits_{j = 1}^{\frac{b}{d}}[\gcd(i, j) = 1]\\ = \sum\limits_{i = 1}^{\frac{a}{d}}\sum\limits_{j = 1}^{\frac{b}{d}}\sum\limits_{e | i, j}\mu(e)\\ = \sum\limits_{e = 1}^{\min(\frac{a}{d}, \frac{b}{d})}\mu(e)\sum\limits_{i = 1}^{\frac{a}{ed}}\sum\limits_{j = 1}^{\frac{b}{ed}}1\\ = \sum\limits_{e = 1}^{\min(\frac{a}{d}, \frac{b}{d})}\mu(e)\lfloor\frac{a}{ed}\rfloor\lfloor\frac{b}{ed}\rfloor\\ $$

直接数论分块即可

4|0能量采集

原问题即求

$$(2 \times\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}\gcd(i, j)) - nm$$

其中

$$\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}\gcd(i, j)\\ = \sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}\sum\limits_{k = 1}^{\min(n, m)}k[\gcd(i, j) = k]\\ = \sum\limits_{k = 1}^{\min(n, m)}k \times\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}[\gcd(i, j) = k]\\ = \sum\limits_{k = 1}^{\min(n, m)}k \times\sum\limits_{i = 1}^{\frac{n}{k}}\sum\limits_{j = 1}^{\frac{m}{k}}[\gcd(i, j) = 1]\\ = \sum\limits_{k = 1}^{\min(n, m)}k \times\sum\limits_{i = 1}^{\frac{n}{k}}\sum\limits_{j = 1}^{\frac{m}{k}}\sum\limits_{e | i, j} \mu(e)\\ = \sum\limits_{k = 1}^{\min(n, m)}k\times\sum\limits_{e = 1}^{\frac{\min(n, m)}{k}}\mu(e)\sum\limits_{i = 1}^{\frac{n}{ke}}\sum\limits_{j = 1}^{\frac{m}{ke}}1\\ = \sum\limits_{k = 1}^{\min(n, m)}k\times\sum\limits_{e = 1}^{\frac{\min(n, m)}{k}}\mu(e)\lfloor\frac{n}{ke}\rfloor\lfloor\frac{m}{ke}\rfloor\\ $$

数论分块套数论分块即可

5|0LCMs

令 $\operatorname{number}(i) = \sum_{j = 0}^{n - 1} [A_j = i]$ ， $m = \max_{j = 0}^{n - 1}(A_j)$

$$\sum\limits_{i=0}^{n-2}\ \sum\limits_{j=i+1}^{n-1}\ \mathrm{lcm}(A_i,A_j)\\ = \frac{(\sum\limits_{i = 1}^{m}\sum\limits_{j = 1}^{m}\mathrm{lcm}(i, j)\times\mathrm{number}(i)\times\mathrm{number}(j)) - \sum\limits_{i = 0}^{n - 1}A_i}{2}$$

其中

$$\sum\limits_{i = 1}^{m}\sum\limits_{j = 1}^{m}\mathrm{lcm}(i, j)\times\mathrm{number}(i)\times\mathrm{number}(j)\\ =\sum\limits_{i = 1}^{m}\sum\limits_{j = 1}^{m}\mathrm{number}(i)\times\mathrm{number}(j)\times\frac{ij}{\gcd(i, j)}\\ =\sum\limits_{i = 1}^{m}\sum\limits_{j = 1}^{m}\mathrm{number}(i)\times\mathrm{number}(j)\times\sum\limits_{k}^{m}\frac{ij}{k}[\gcd(i, j) = k]\\ =\sum\limits_{k}^{m}\sum\limits_{i = 1}^{m}\sum\limits_{j = 1}^{m}\mathrm{number}(i)\times\mathrm{number}(j)\times\frac{ij}{k}[\gcd(i, j) = k]\\ =\sum\limits_{k}^{m}k\times\sum\limits_{i = 1}^{\frac{m}{k}}\sum\limits_{j = 1}^{\frac{m}{k}}\mathrm{number}(ik)\times\mathrm{number}(jk)\times ij[\gcd(i, j) = 1]\\ =\sum\limits_{k}^{m}k\times\sum\limits_{i = 1}^{\frac{m}{k}}\sum\limits_{j = 1}^{\frac{m}{k}}\mathrm{number}(ik)\times\mathrm{number}(jk)\times ij\sum\limits_{e | i, j}\mu(e)\\ =\sum\limits_{k}^{m}k\times\sum\limits_{e}^{\frac{m}{k}}\mu(e) \times e^{2}\sum\limits_{i = 1}^{\frac{m}{ke}}\sum\limits_{j = 1}^{\frac{m}{ke}}\mathrm{number}(ike)\times\mathrm{number}(jke)\times ij\\ =\sum\limits_{e}^{m}\mu(e) \times e^{2}\sum\limits_{k}^{\frac{m}{e}}k\times\sum\limits_{i = 1}^{\frac{m}{ke}}\sum\limits_{j = 1}^{\frac{m}{ke}}\mathrm{number}(ike)\times\mathrm{number}(jke)\times ij\\ =\sum\limits_{e}^{m}\mu(e) \times e^{2}\sum\limits_{k}^{\frac{m}{e}}k\times\sum\limits_{i = 1}^{\frac{m}{ke}}\mathrm{number}(ike)\times i\sum\limits_{j = 1}^{\frac{m}{ke}}\mathrm{number}(jke)\times j\\ =\sum\limits_{e}^{m}\mu(e) \times e^{2}\sum\limits_{k}^{\frac{m}{e}}k\times(\sum\limits_{i = 1}^{\frac{m}{ke}}\mathrm{number}(ike)\times i)^{2}\\ $$

对于每个 $ke$ 记录 $\sum\limits_{i = 1}^{\frac{m}{ke}}\mathrm{number}(ike)\times i$ ， 枚举 $e$ 和 $k$， 看似会超时实则调和级数加主定理

6|0数表

考虑没有限制， 则原题为

$$\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}\sum\limits_{k | i, j} k\\ = \sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}\sum\limits_{k | \gcd(i, j)} k\\ = \sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}\sigma(\gcd(i, j))\\ = \sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}\sum\limits_{k = 1}^{\min(n, m)}\sigma(k)[\gcd(i, j) = k]\\ = \sum\limits_{k = 1}^{\min(n, m)} \sum\limits_{i = 1}^{\frac{n}{k}}\sum\limits_{j = 1}^{\frac{m}{k}}\sigma(k)[\gcd(i, j) = 1]\\ = \sum\limits_{k = 1}^{\min(n, m)}\sigma(k) \sum\limits_{i = 1}^{\frac{n}{k}}\sum\limits_{j = 1}^{\frac{m}{k}}\sum\limits_{e | i, j}\mu(e)\\ = \sum\limits_{k = 1}^{\min(n, m)}\sigma(k)\sum\limits_{e = 1}^{\min(\frac{n}{k}, \frac{m}{k})}\mu(e) \sum\limits_{i = 1}^{\frac{n}{ke}}\sum\limits_{j = 1}^{\frac{m}{ke}}1\\ \sum\limits_{k = 1}^{\min(n, m)}\sigma(k)\sum\limits_{e = 1}^{\min(\frac{n}{k}, \frac{m}{k})}\mu(e) \lfloor\frac{n}{ke}\rfloor\lfloor\frac{m}{ke}\rfloor\\ $$

但这个样子对于不同的 $n$ ， $m$ 至少要用 $\mathrm{O}(n\sqrt{n})$ 的时间求解， 不满足多组询问的要求， 所以考虑变换， 将常数项往后提

$$\therefore \sum\limits_{k = 1}^{\min(n, m)}\sigma(k)\sum\limits_{e = 1}^{\min(\frac{n}{k}, \frac{m}{k})}\mu(e) \lfloor\frac{n}{ke}\rfloor\lfloor\frac{m}{ke}\rfloor\\ = \sum\limits_{i}^{min (n, m)}\lfloor\frac{n}{i}\rfloor\lfloor\frac{m}{i}\rfloor \sum\limits_{k | i}\sigma(k)\mu(\frac{i}{k})\\ $$

这里可以考虑对不同的 $i$ 维护 $\sum\limits_{k | i}\sigma(k)\mu(\frac{i}{k})$ ， 并对 $i$ 求前缀和， 这样对于一次操作我们就可以用 $\mathrm{O}(\sqrt{n})$ 的时间复杂度去求解。 再考虑题目限制， 即要求 $\sigma(k)$ 不能超过 $a$ ， 那么我们考虑对所有会参与计算的 $\sigma(k)\mu(\frac{i}{k})$ 排序， 因为调和级数， 所以这里时间复杂度是 $O(n\log^{2}{n})$ 的， 然后再对询问的 $a$ 排序， 那么每次只会增加一些 $\sigma(k)\mu(\frac{i}{k})$ 可以产生贡献， 我们把刚才的前缀和改为 BIT ， 每次就在上面修改和查询即可， 时间复杂度 $O(q\sqrt{n}\log{n} + n\log^{2}{n})$

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本文作者：FlowerDream
本文链接：https://www.cnblogs.com/flower-dream/p/17736506.html
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