51nod - 1363 - 最小公倍数之和 - 数论

https://www.51nod.com/Challenge/Problem.html#!#problemId=1363
求\(\sum\limits_{i=1}^{n}lcm(i,n)\)
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先换成gcd
\(\sum\limits_{i=1}^{n}\frac{i*n}{gcd(i,n)}\)

显而易见，枚举g
$ n * \sum\limits_{g|n} \frac{1}{g} \sum\limits_{i=1}^{n} i*[gcd(i,n)==g] $

提g，没有下整符号
$ n * \sum\limits_{g|n} \frac{1}{g} \sum\limits_{i=1}^{\frac{n}{g}} i*[gcd(i,\frac{n}{g})==1] $

标签：frac,gcd,limits,公倍数,sum,51nod,1363
来源： https://www.cnblogs.com/Yinku/p/10987912.html