huntian oy

https://acm.hdu.edu.cn/showproblem.php?pid=6706
求

\[\sum_{i=1}^n\sum_{j=1}^igcd(i^m-j^m, i^n-j^n)[(i, j)=1] \]

已知公式

\[a > b, gcd(a^m-b^m, a^n-b^n) = a^{gcd(m, n)}-b^{gcd(m,n)} \]

题目保证 \([(n, m) = 1]\) 那么转化为

\[\sum_{i=1}^n\sum_{j=1}^i i- j[(i, j)=1]\\ =\sum_{i=1}^n(\sum_{j=1}^ii[(i, j)=1] - \sum_{j=1}^nj[(i, j) = 1])\\ =\sum_{i=1}^n(i*φ(i)-\frac {i*φ(i) + [i = 1]} 2)\\ =\sum_{i=1}^n\frac {i * φ(i) - [i = 1]} 2\\ 2*f(n)+1 = \sum_{i=1}^ni*φ(i) \]