失踪人口回归
根据\(Polya\)定理\[ans=\frac 1n \sum\limits_{i=1}^n2^{gcd(i, n)}\]
考虑枚举\(gcd\)，原式变成\[\frac 1n \sum\limits_{d|n}2^d\sum\limits_{i=1}^n\big[gcd(i,n)=d\big]\]
用\(id\)替换\(i\) \[\frac 1n \sum\limits_{d|n}2^d\sum\limits_{i=1}^{\frac nd}\big[gcd(i,\frac nd)=1\big]\]
\(\sum\limits_{i=1}^{\frac nd}\big[gcd(i,\frac nd)=1\big]\)这个东西显然是\(\varphi(\frac nd)\)，所以答案即为\[\frac 1n \sum\limits_{d|n}2^d\varphi(\frac nd)\]
这题比较恶心的地方就是高精