数学大佬%%%
by chen_zhe @ 2017-10-06 16:54:57
@[Js2xxx](/space/show?uid=49189) 会超时啊
by xw001 @ 2017-10-26 08:25:19
@[xw001](/space/show?uid=15044) 不会啊
by BlueArc @ 2017-12-30 22:20:18
只需要二分出导函数的零点即可
by aiyougege @ 2018-07-07 17:33:06
orzorz%%
by dyx131313 @ 2018-08-09 10:57:15
牛顿迭代法
by A星际穿越 @ 2018-08-21 14:25:23
orz
tql
%%%
by 13802919466djh @ 2022-08-03 09:53:32