```cpp
g(0)=0,g(1)=1
g(n)=[2002*g(n-1)+2003*g(n-2)]%2005
=[(-3)*g(n-1)+(-2)*g(n-2)]%2005
则:g(n)+g(n-1)=[(-2)*(g(n-1)+g(n-2))]
```
by lxzy_ @ 2020-04-01 23:23:07
```cpp
g(0)=0,g(1)=1
g(n)=[2002*g(n-1)+2003*g(n-2)]%2005
=[(-3)*g(n-1)+(-2)*g(n-2)]%2005
则:g(n)+g(n-1)=[(-2)*(g(n-1)+g(n-2))]%2005
```
by 250250250sbsbsb @ 2020-04-01 23:28:12
哦对是这样的,谢谢
by lxzy_ @ 2020-04-01 23:30:50
各位大佬在吗?
@[蒟蒻的名字](/user/147999)
@Victory_Deafeat
@[数学王子](/user/201493)
@各位大佬
by lxzy_ @ 2020-04-01 23:31:57
@[Victory_Defeat](/user/70592)
by lxzy_ @ 2020-04-01 23:32:41
@[chen__zhe](/user/128417)
by do_while_false @ 2020-04-01 23:37:35
@[流星之愿](/user/67493) 大佬还有:function_of_zero、Alpha、void_basic_learner
by BlueSu @ 2020-04-01 23:38:22
@[流星之愿](/user/67493) 同余(应该是)
by Steven__Chen @ 2020-04-01 23:45:19
就是把g[n+1]拿了一个出来吧
和同余有啥关系
by SSerxhs @ 2020-04-02 00:08:18
### 我觉得这个应该是同余吧,如果不是请无视我这条
因为 $g_n=\big[(-3)\cdot g_{n-1}+(-2)\cdot g_{n-2} \big] \operatorname{mod} 2005 $
所以:
$$g_n+g_{n-1}=\big[(-3)\cdot g_{n-1}+(-2)\cdot g_{n-2} \big] \bmod 2005 + g_{n-1}$$
$$\equiv\big[(-3)\cdot g_{n-1}+(-2)\cdot g_{n-2} +g_{n-1}\big] (\bmod\ 2005)$$
$$\equiv\big[(-2)\cdot g_{n-1}+(-2)\cdot g_{n-2} \big] (\bmod\ 2005)$$
$$\equiv\big[(-2)\cdot (g_{n-1}+ g_{n-2}) \big] (\bmod\ 2005)$$
by 江户川·萝卜 @ 2020-04-02 07:28:49