直接取模啊
by Jasonying @ 2020-07-17 20:31:00
@[shimao](/user/312121)
$(a+b) \operatorname{mod} c=a\operatorname{mod} c+b\operatorname{mod} c$
$(a*b) \operatorname{mod} c=a\operatorname{mod} c*b\operatorname{mod} c$
by Smile_Cindy @ 2020-07-17 20:31:05
@[shimao](/user/312121) 建议去搜“同余定理”
by 一只书虫仔 @ 2020-07-17 20:31:18
@[Alpha](/user/87058) ~~右边还要再模一次吧（~~
by _5011_ @ 2020-07-17 20:31:36
@[Alpha](/user/87058) 不对吧 应该还要mod一遍吧
by Jasonying @ 2020-07-17 20:31:50
那a-b、a/b呢
by shimao @ 2020-07-17 20:35:36
@[Alpha](/user/87058)
by shimao @ 2020-07-17 20:35:41
@[shimao](/user/312121) a-b同理，a/b得用逆元
by _5011_ @ 2020-07-17 20:37:03
@[shimao](/user/312121)
$$
\begin{matrix}
(a+b)\bmod c=a\bmod c+b\bmod c\\
(a\times b)\bmod c=(a\bmod c)\times(b\bmod c)
\end{matrix}
$$
by CSP_Sept @ 2020-07-17 20:39:00
即，对于 $a/b$，你需要使用一些手段求出满足 $xb\equiv 1\pmod p$ 的 $x$，并用 $a$ 乘 $x$。
by _5011_ @ 2020-07-17 20:39:05