矩阵乘法-洛谷P2233 [HNOI2002] 公交车路线

https://daniu.luogu.org/problem/show?pid=2233
矩阵这个很显然啊；
然后直接快速幂就好了；
至于为什么，这个就是矩阵的基本性质；
可以看我相关的博客；
然后到了E点就不懂，直接在乘法的时候处理一下就好了

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<map>
#define Ll long long
using namespace std;
int a[9][9]
{//A B C D E F G H
{0,0,0,0,0,0,0,0,0},
{0,0,1,0,0,0,0,0,1},//A
{0,1,0,1,0,0,0,0,0},//B
{0,0,1,0,1,0,0,0,0},//C
{0,0,0,1,0,1,0,0,0},//D
{0,0,0,0,1,0,1,0,0},//E
{0,0,0,0,0,1,0,1,0},//F
{0,0,0,0,0,0,1,0,1},//G
{0,1,0,0,0,0,0,1,0},//H
};
struct jv{
    int m[9][9];
    jv(){memset(m,0,sizeof m);}
}ans,c;
int n,mo=1e3;
jv cheng(jv x,jv y){
    jv z;
    for(int i=1;i<=8;i++)
    for(int j=1;j<=8;j++)
    for(int k=1;k<=8;k++)if(k!=5)
    z.m[i][j]=(z.m[i][j]+x.m[i][k]*y.m[k][j])%mo;
    return z;
}
int main()
{
    scanf("%d",&n);
    n--;
    for(int i=1;i<=8;i++)
    for(int j=1;j<=8;j++)
        c.m[i][j]=a[i][j];
    ans=c;
    while(n){
        if(n&1)ans=cheng(ans,c);
        n>>=1;
        c=cheng(c,c);
    }
    printf("%d",ans.m[1][5]);
}