dalao,我问一下这道题,给定数列
1
,
1
,
1
,
3
,
5
,
9
,
17
,
⋯
从第
4 项开始,每项都是前
3 项的和。求第
20190324项的最后
4位数字。
我每加一次,都要mod10000,为啥mod10000之后不影响结果呢
by LiChenyi05 @ 2024-04-02 22:08:06
@[Tomle](/user/942161)
by LiChenyi05 @ 2024-04-02 22:08:36
因为 $(a+b)\operatorname{mod}p=(a\operatorname{mod}p+b\operatorname{mod}p)\operatorname{mod}p$,也就是说,两数相加取余等于两数取余再相加再取余,这是取余的基本性质
by Tomle @ 2024-04-03 06:51:50
@[LiChenyi05](/user/1269314)
by Tomle @ 2024-04-03 06:52:10
减法也满足,但要特判一下负数,$(a-b)\operatorname{mod}p=(a\operatorname{mod}p-b\operatorname{mod}p+p)\operatorname{mod}p$
乘法也满足,$(a \times b) \operatorname{mod}p=(a \operatorname{mod}p)\times(b \operatorname{mod}p)\operatorname{mod}p$
但是除法不行,除法取余是数论里的知识了
by Tomle @ 2024-04-03 06:58:35
幂也能取余 $a^b\operatorname{mod}p=(a\operatorname{mod}p)^{b\operatorname{mod}p}\operatorname{mod}p$
所以,加减乘幂能在运算中随时取余,不影响结果
by Tomle @ 2024-04-03 07:02:22
我错了,$a^b\operatorname{mod}=(a\operatorname{mod}p)^b\operatorname{mod} p$
by Tomle @ 2024-04-03 07:09:37
@[Tomle](/user/942161) 谢谢您
by LiChenyi05 @ 2024-04-03 08:13:26