24K卡布奇诺 - 2023/2/27 16:13:00
给前面提到的第一种算法加上MIT许可证：
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见崎鸣 - 2023/3/29 21:28:00
末学认为4楼提出的假设可以细分成下图的情况，当然讨论的[b]前提[/b]是有潜水的玩家。
[img=637,486]https://www.keyfc.net/bbs/space/upload/2023/03/29/1030934211.png[/img]
可以看出“[b]奇偶是否相同[/b]”正好就对应了“[b]是否可以排水[/b]”。末学的理解是这样的：当同奇或同偶时，排水看似减少了一些平民赢的情况，但是更多减少的是狼人赢的情况，比如一些在投票前就已经结束的情况。“伤敌一千，自损八百”了属于是。
而当一奇一偶时，排水难以减少狼人赢的情况，甚至会增加，所以此时排水不可行。
:miffy5:
末学还想改进一下算法，附近至少要观察多少人。不会插入表格，那就用csv格式吧。
,,天数n,,,,
,b_mn,1,2,3,4,
m,1,0.637,-0.318,0.212,-0.159,
,2,-0.273,0.318,-0.141,0.0795,
,3,0.463,-0.664,0.953,-0.511,
,4,-0.854,0.328,-0.604,0.511,
,5,0.743,-0.688,0.331,-0.614,
列表长度a，场上杀手数e，a mod e=m，偏移量o=[a*b]（向下取整）,,,,,,
"至少观察t人，使C(a,t)>5C(a-e,t)",,,,,,
a,e,m,n,b,o,
19,5,4,1,-0.854,-17,
,,,,,,
1,水羊,,,,,
2,陆陆侠,,,,,
3,sakyas,,,,,
4,来自宇宙的乱码兄,,,,,
5,sxqsxq,,,,,
6,列克星敦太太,,,,,
7,夏凪緋雪,,,,,
8,ako,,,,,
9,Citrus,,,,,
10,红叶晚潇潇,,,,,
11,鬼烎,,,,,
12,Key_Player,,,,,
13,莉莉,,,,,
14,24K卡布奇诺,,,附近2,,
15,哈利路亚100晴人,观察者,15-17+19=17,,,
16,amlut,,附近1,,,
17,nemoma,被观察,"C(19,5)>5C(14,5),t=5",,,
18,Inle,,附近1,,,
19,0plume0,,,附近2,,