专栏文章

题解:P8570 [JRKSJ R6] 牵连的世界

P8570题解参与者 1已保存评论 0

文章操作

快速查看文章及其快照的属性,并进行相关操作。

当前评论
0 条
当前快照
1 份
快照标识符
@mine2pge
此快照首次捕获于
2025/12/02 00:54
3 个月前
此快照最后确认于
2025/12/02 00:54
3 个月前
查看原文
i=1nj=1mc(ij)φ(ij)=i=1nj=1mφ(i)φ(j)gcd(i,j)φ(gcd(i,j))xiyi[gcd(x,y)=1]=i=1nj=1mφ(i)φ(j)gcd(i,j)φ(gcd(i,j))xiyidx,dyμ(d)=i=1nj=1mφ(i)φ(j)gcd(i,j)φ(gcd(i,j))d=1c(id)c(jd)μ(d)=d=1μ(d)i=1ndj=1mdφ(id)φ(jd)gcd(i,j)dφ(gcd(i,j)d)c(i)c(j)=d=1μ(d)dk=1i=1ndkj=1mdkφ(idk)φ(jdk)[gcd(i,j)=1]kφ(kd)c(ik)c(jk)=d=1μ(d)dk=1kφ(kd)i=1ndkφ(idk)c(ik)j=1mdkφ(jdk)c(jk)wi,wjμ(w)=d=1μ(d)dk=1kφ(kd)w=1μ(w)i=1ndkwφ(idkw)c(ikw)j=1mdkwφ(jdkw)c(jkw)\sum_{i=1}^n\sum_{j=1}^mc(ij)\varphi(ij)\\ =\sum_{i=1}^n\sum_{j=1}^m\frac{\varphi(i)\varphi(j)\gcd(i,j)}{\varphi(\gcd(i,j))}\sum_{x|i}\sum_{y|i}[\gcd(x,y)=1]\\ =\sum_{i=1}^n\sum_{j=1}^m\frac{\varphi(i)\varphi(j)\gcd(i,j)}{\varphi(\gcd(i,j))}\sum_{x|i}\sum_{y|i}\sum_{d|x,d|y}\mu(d)\\ =\sum_{i=1}^n\sum_{j=1}^m\frac{\varphi(i)\varphi(j)\gcd(i,j)}{\varphi(\gcd(i,j))}\sum_{d=1}c(\frac id)c(\frac jd)\mu(d)\\ =\sum_{d=1}\mu(d)\sum_{i=1}^{\frac nd}\sum_{j=1}^{\frac md}\frac{\varphi(id)\varphi(jd)\gcd(i,j)d}{\varphi(\gcd(i,j)d)}c(i)c(j)\\ =\sum_{d=1}\mu(d)d\sum_{k=1}\sum_{i=1}^{\frac n{dk}}\sum_{j=1}^{\frac m{dk}}\frac{\varphi(idk)\varphi(jdk)[\gcd(i,j)=1]k}{\varphi(kd)}c(ik)c(jk)\\ =\sum_{d=1}\mu(d)d\sum_{k=1}\frac k{\varphi(kd)}\sum_{i=1}^{\frac n{dk}}\varphi(idk)c(ik)\sum_{j=1}^{\frac m{dk}}\varphi(jdk)c(jk)\sum_{w|i,w|j}\mu(w)\\ =\sum_{d=1}\mu(d)d\sum_{k=1}\frac k{\varphi(kd)}\sum_{w=1}\mu(w)\sum_{i=1}^{\frac n{dkw}}\varphi(idkw)c(ikw)\sum_{j=1}^{\frac m{dkw}}\varphi(jdkw)c(jkw)\\

评论

0 条评论,欢迎与作者交流。

正在加载评论...