n在边输入边取模就可以了。
by Ether13 @ 2024-04-12 20:45:57
$$
\begin{aligned}
\text{原式}&=(\sum_{i=1}^{m}i\cdot \left\lfloor\dfrac{n}{m}\right\rfloor+\sum_{i=1}^{n\bmod m}i)\bmod m\\
\text{令 } t&=n\bmod m\\
\text{原式}&=(\sum_{i=1}^{m}i\cdot \dfrac{n-t}{m}+\sum_{i=1}^{t}i)\bmod m\\
&=(\dfrac{m(m+1)}{2}\cdot\dfrac{n-t}{m}+\dfrac{t(t+1)}{2})\bmod m\\
&=(\dfrac{(m+1)(n-t)+t(t+1)}{2})\bmod m
\end{aligned}
$$
后面就不会了(bushi),但是可能会给你一丢丢帮助
by jqQt0220 @ 2024-04-12 20:54:04
具体来说,就是
$$
\frac{(n+1)n}{2}\bmod m=\frac{((n\bmod m)+1)(n\bmod m)}{2}\bmod m
$$
by Ether13 @ 2024-04-12 20:55:25
@[jqQt0220](/user/678175) @[Ether13](/user/515717) 谢谢大佬帮助,好人一生平安。
by ikun_god @ 2024-04-12 20:56:33