The Preliminary Contest for ICPC Asia Nanjing 2019 C. Tsy's number 5

https://nanti.jisuanke.com/t/41300 題意:求\(\sum_{i=1}^n\phi(i)\phi(j)2^{\phi(i)\phi(j)}\) \(f_i=\sum_{k=1}^n[\phi(k)==i]\) \(\sum_{i=1}^n\phi(i)\phi(j)2^{\phi(i)\phi(j)}\) \(=\sum_{i=1}^n\sum_{j=1}^nf_i

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