UESTC - 1610 递推方程+矩阵快速幂

感觉像是HDU Keyboard的加强版，先推出3张牌时的所有组合，然后递推出n张牌

看到n=1e18时吓尿了

最后24那里还是推错了..

(5行1列 dp[1][n],dp[2][n],dp[3][n],dp[4][n],dp[5][n]) = A^(n-3) * (5行1列 4,12,12,12,24)

其中,A=

1,0,0,1,0

3,0,0,3,0

0,1,1,0,1

0,1,1,0,1

0,2,2,0,1

#include<bits/stdc++.h>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
using namespace std;
typedef long long ll;
const int maxn  = 100;
const ll MOD = 1e9+9;
inline ll mod(ll a){return a%MOD;}
int b[6][6]={
    {0,0,0,0,0,0},
    {0,1,0,0,1,0},
    {0,3,0,0,3,0},
    {0,0,1,1,0,1},
    {0,0,1,1,0,1},
    {0,0,2,2,0,1},
};
ll c[6]={0,4,12,12,12,24};
struct Mat{
    ll m[7][7],r,c;
    void node(int rr,int cc,bool unit=0){
        r=rr;c=cc;
        memset(m,0,sizeof m);
        if(unit) rep(i,1,rr) m[i][i]=1;
    }
};
Mat operator * (Mat a,Mat b){
    Mat ans;ans.node(a.r,b.c);
    rep(i,1,a.r){
        rep(j,1,b.c){
            int t=max(a.r,b.c);
            rep(k,1,t){
                ans.m[i][j]+=mod(a.m[i][k]*b.m[k][j]);
                ans.m[i][j]=mod(ans.m[i][j]);
            }
        }
    }
    return ans;
}
Mat qmod(Mat a,ll n){
    Mat ans;ans.node(5,5,1);
    while(n){
        if(n&1) ans=ans*a;
        n>>=1;
        a=a*a;
    }
    return ans;
}
ll qmod(ll a,ll n){
    ll ans=1;
    while(n){
        if(n&1) ans=mod(mod(ans)*mod(a));
        n>>=1;
        a=mod(mod(a)*mod(a));
    }
    return mod(ans);
}
int main(){
    ll n;
    Mat base,base2; base.node(5,5); base2.node(5,1);
    rep(i,1,5) rep(j,1,5) base.m[i][j]=b[i][j];
    rep(i,1,5) base2.m[i][1]=c[i];
    while(scanf("%lld",&n)!=EOF){
        Mat tmp=qmod(base,n-3);
        Mat res=tmp*base2;
        ll ans=0;
        rep(i,1,5) {ans+=mod(res.m[i][1]);ans=mod(ans);}
        ans=mod(qmod(4,n)-ans);
        printf("%lld\n",ans<0?ans+MOD:ans);
    }
    return 0;
}