D - Covering
Bob's school has a big playground, boys and girls always play games here after school.
To protect boys and girls from getting hurt when playing happily on the playground, rich boy Bob decided to cover the playground using his carpets.
Meanwhile, Bob is a mean boy, so he acquired that his carpets can not overlap one cell twice or more.
He has infinite carpets with sizes of 1×21×2 and 2×12×1, and the size of the playground is 4×n4×n.
Can you tell Bob the total number of schemes where the carpets can cover the playground completely without overlapping?
Input
There are no more than 5000 test cases.
Each test case only contains one positive integer n in a line.
1≤n≤10181≤n≤1018
Output
For each test cases, output the answer mod 1000000007 in a line.
Sample Input
1 2
Sample Output
1 5
//递推公式黑科技
#include<bits/stdc++.h>
using namespace std;
///#define X first
//#define Y second
#define PB push_back
//#define MP make_pair
//#define MEM(x,y) memset(x,y,sizeof(x));
//#define bug(x) cout<<"bug"<<x<<endl;
typedef long long ll;
//typedef pair<int,int> pii;
using namespace std;
//const int maxn=1e3+10;
const int mod=1000000007; //按实际改
ll powmod(ll a,ll b){ //快速幂
ll res=1;a%=mod;
assert(b>=0);
while(b){
if(b&1) res=res*a%mod; a=a*a%mod; b>>=1;
}
return res;
}
// head
namespace linear_seq {
const int N=10010; //不需改
ll res[N],base[N],_c[N],_md[N];
vector<ll> Md;
void mul(ll *a,ll *b,int k) {
for(int i=0;i<k+k;i++) _c[i]=0;
for(int i=0;i<k;i++) if (a[i])
for(int j=0;j<k;j++) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
for (ll i=k+k-1;i>=k;i--) if (_c[i])
for(int j=0;j<Md.size();j++)
_c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
for(int i=0;i<k;i++) a[i]=_c[i];
}
int solve(ll n,vector<ll> a,vector<ll> b) {
// a 系数 b 初值 b[n+1]=a[0]*b[n]+...
//求出的是第n+1项
ll ans=0,pnt=0;
ll k=a.size();
assert(a.size()==b.size());
for(int i=0;i<k;i++) _md[k-1-i]=-a[i];_md[k]=1;
Md.clear();
for(int i=0;i<k;i++) if (_md[i]!=0) Md.push_back(i);
for(int i=0;i<k;i++) res[i]=base[i]=0;
res[0]=1;
while ((1ll<<pnt)<=n) pnt++;
for (ll p=pnt;p>=0;p--) {
mul(res,res,k);
if ((n>>p)&1) {
for (ll i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
for(int j=0;j<Md.size();j++) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
}
}
for(int i=0;i<k;i++) ans=(ans+res[i]*b[i])%mod;
if (ans<0) ans+=mod;
return ans;
}
vector<ll> BM(vector<ll> s) {
vector<ll> C(1,1),B(1,1);
int L=0,m=1,b=1;
for(int n=0;n<s.size();n++) {
ll d=0;
for(int i=0;i<L+1;i++) d=(d+(ll)C[i]*s[n-i])%mod;
if (d==0) ++m;
else if (2*L<=n) {
vector<ll> T=C;
ll c=mod-d*powmod(b,mod-2)%mod;
while (C.size()<B.size()+m) C.PB(0);
for(int i=0;i<B.size();i++) C[i+m]=(C[i+m]+c*B[i])%mod;
L=n+1-L; B=T; b=d; m=1;
} else {
ll c=mod-d*powmod(b,mod-2)%mod;
while (C.size()<B.size()+m) C.PB(0);
for(int i=0;i<B.size();i++) C[i+m]=(C[i+m]+c*B[i])%mod;
++m;
}
}
return C;
}
int gao(vector<ll> a,ll n) {
vector<ll> c=BM(a);
c.erase(c.begin());
for(int i=0;i<c.size();i++) c[i]=(mod-c[i])%mod;
return solve(n,c,vector<ll>(a.begin(),a.begin()+c.size()));
}
};
//用的时候只用改mod的值和前几项的数值
int main(){
ll n;
while(~scanf ("%lld", &n)){ //求第n项 //一般放入前8项
printf("%lld\n",linear_seq::gao(vector<ll>{1,5,11,36,95,281,781,2245},n-1));
}
}