感覺像是HDU Keyboard的加強(qiáng)版，先推出3張牌時的所有組合，然后遞推出n張牌
看到n=1e18時嚇尿了
最后24那里還是推錯了..
(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;
}
轉(zhuǎn)載于:https://www.cnblogs.com/caturra/p/8310605.html
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