`f[i]` i变为n的期望步数
`f[n]=0`
`f[i]=p*f[i+1]+(1-p)*f[0]+1`
答案为 `f[0]`
by LHQing @ 2020-08-06 08:14:33
楼上正解(
by Polaris_Dane @ 2020-08-06 08:16:58
@[LHQing](/user/167507) 请问要这东西是要高斯消元吗/kk
by zhoukangyang @ 2020-08-06 08:20:41
还是有更快的方法qwq
by zhoukangyang @ 2020-08-06 08:21:02
@[zhoukangyang](/user/173660) 不用吧,这个可以直接看成跟$f[0]$相关的方程式,递推他的系数
by Polaris_Dane @ 2020-08-06 08:23:24
能不能考虑保存
```
f[i] = k f[0] + b
```
by 王鲲鹏 @ 2020-08-06 08:24:05
啊好像是的, 谢谢
by zhoukangyang @ 2020-08-06 08:27:43
@[LHQing](/user/167507) 听说答案是(1/p)^n
by zzqDeco @ 2020-08-06 08:29:12
可以快速推出吗
by zzqDeco @ 2020-08-06 08:29:43
歌唱王国?(
by wurzang @ 2020-08-06 09:07:09