$$\sum_{i=1}^n\sum_{j=1}^ngcd(i,j)$$
by chihik @ 2019-12-12 13:18:23
@[等风吹来](/user/123802)
$ Ans(n)=\sum_{i=1}^n\sum_{j=1}^ngcd(i,j) $
by nkinclude @ 2019-12-12 13:23:45
@[等风吹来](/user/123802)
$Ans(n)=Ans(n-1)+***$
by nkinclude @ 2019-12-12 13:24:07
这个不算吧
by Curators @ 2019-12-12 13:25:07
@[DICalculus](/user/86409) ？
by nkinclude @ 2019-12-12 13:26:06
@[nkinclude](/user/186182) 怎么递推计算?
by chihik @ 2019-12-12 13:27:42
？？？
by nkinclude @ 2019-12-12 13:28:01
@[等风吹来](/user/123802) 似乎不用那个带根号的式子
by nkinclude @ 2019-12-12 13:29:47
@[等风吹来](/user/123802) 差分后可以直接筛出来
by hellomath @ 2019-12-12 13:33:12
@[等风吹来](/user/123802) 话说这是哪道题啊
by nkinclude @ 2019-12-12 13:48:24