@[chen_zhe](/space/show?uid=8457)
by Fading @ 2018-09-14 13:57:33
$X_i+1≡aX_i+b(mod$ $p)$
规范一点的
by Fading @ 2018-09-14 14:01:43
```
$ X_{i+1}=aX_i+b (mod$ $p)$
```
$ X_{i+1}=aX_i+b (mod$ $p)$
by jzzcjb @ 2018-09-14 14:18:45
$ X_{i+1}=aX_i+b (mod$ $p)$
by Fading @ 2018-09-14 14:20:24
@[jzzcjb](/space/show?uid=57304) @[我是一个菜鸡](/space/show?uid=20309)
`\pmod` 了解一下
```plain
$ X_{i+1}=aX_i+b \pmod p$
```
$ X_{i+1}=aX_i+b \pmod p$
by Planet6174 @ 2018-09-14 14:23:28
@[yjjr](/space/show?uid=5088) @[龟龟号打捞船](/space/show?uid=36482)
by Planet6174 @ 2018-09-14 14:23:52
@[jzzcjb](/space/show?uid=57304)
正解：
$X_{i+1} \equiv aX_i+b \pmod p$
```
$X_{i+1} \equiv aX_i+b \pmod p$
```
by 幸济蒸馒头 @ 2018-09-14 14:25:12
@[Planet6174](/space/show?uid=29762) 呀，我正想说pmod
by 幸济蒸馒头 @ 2018-09-14 14:25:41
蜜汁尴尬
by 幸济蒸馒头 @ 2018-09-14 14:26:20
@[我是一个菜鸡](/space/show?uid=20309)
@[jzzcjb](/space/show?uid=57304)
@[Planet6174](/space/show?uid=29762)
@[RE自动机](/space/show?uid=80898)
感谢您们的贡献
by chen_zhe @ 2018-09-14 20:01:02