p4449-solution

P4449 Solution

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\[\begin{aligned} \sum_{i=1}^n\sum_{j=1}^m\gcd(i,j)^k&=\sum_{d=1}^n\sum_{i=1}^n\sum_{j=1}^md^k[\gcd(i,j)=d]\\ &=\sum_{d=1}^n\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}d^k[\gcd(i,j)=1]\\ &=\sum_{d=1}^n\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}d^k\sum_{s|i,s|j}\mu(s)\\ &=\sum_{d=1}^nd^k\sum_{s=1}^{n/d}\mu(s)\lfloor\frac n{ds}\rfloor\lfloor\frac m{ds}\rfloor\\ &=\sum_{t=1}^n\lfloor\frac n t\rfloor\lfloor\frac m t\rfloor\sum_{d|t}d^k\mu(\frac t d) \end{aligned}\]

令 \(f(n)=\sum_{d|n}d^k\mu(\frac n d)\)，这个显然可以线性筛

具体地，对于质数 \(p\)，\(f(p)=p^k-1\)，\(f(p^r)=p^{kr}-p^{(k-1)r}\)，即 \(f(p^r)=p^kf(p^{r-1})\)

线性筛再分块即可。