#include <bits/stdc++.h> using namespace std; #define rep(i,a,n) for (long long i=a;i<n;i++) #define per(i,a,n) for (long long i=n-1;i>=a;i--) #define pb push_back #define mp make_pair #define all(x) (x).begin(),(x).end() #define fi first #define se second #define SZ(x) ((long long)(x).size()) typedef vector<long long> VI; typedef long long ll; typedef pair<long long,long long> PII; const ll mod=1e9+7; ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} // head long long _,n; namespace linear_seq { const long long N=10010; ll res[N],base[N],_c[N],_md[N]; vector<long long> Md; void mul(ll *a,ll *b,long long k) { rep(i,0,k+k) _c[i]=0; rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod; for (long long i=k+k-1;i>=k;i--) if (_c[i]) rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod; rep(i,0,k) a[i]=_c[i]; } long long solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+... // printf("%d\n",SZ(b)); ll ans=0,pnt=0; long long k=SZ(a); assert(SZ(a)==SZ(b)); rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1; Md.clear(); rep(i,0,k) if (_md[i]!=0) Md.push_back(i); rep(i,0,k) res[i]=base[i]=0; res[0]=1; while ((1ll<<pnt)<=n) pnt++; for (long long p=pnt;p>=0;p--) { mul(res,res,k); if ((n>>p)&1) { for (long long i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0; rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod; } } Receives (and 0 , k) = years (years + nothing [i] * b [i])% mod; f (years < 0 ) ans = + mod; return years; } VI BM (VI s) { VI C ( 1 , 1 ), B ( 1 , 1 ); long length L = 0 , m = 1 , b = 1 ; receives (n, 0 , SZ (s)) { ll d = 0 ; receives (and 0,L+1) d=(d+(ll)C[i]*s[n-i])%mod; if (d==0) ++m; else if (2*L<=n) { VI T=C; ll c=mod-d*powmod(b,mod-2)%mod; while (SZ(C)<SZ(B)+m) C.pb(0); rep(i,0,SZ(B)) 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 (SZ(C)<SZ(B)+m) C.pb(0); rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod; ++m; } } return C; } long long gao(VI a,ll n) { VI c=BM(a); c.erase(c.begin()); rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod; return solve(n,c,VI(a.begin(),a.begin()+SZ (C))); } }; int main () { the while (~ Scanf ( " % I64d " , & n)) { / * find the n-th item * / the printf ( " % I64d \ n " , linear_seq :: Gao ({Vl . 1 , . 5 , . 11 , 36 , 95 , 281 is , 781 , 2245 , 6336 , 18061 , 51205 }, N- . 1 )); / * output coefficients * / / * first k entries recursion need 2 * k items can be determined*/ VI res = linear_seq::BM(VI{1,5,11,36,95,281,781,2245,6336,18061, 51205}); for(int i = 1;i < res.size();i++) printf("%lld\n", (mod-res[i]) % mod); //f(n) = f(n-1) + 5*f(n-2) + f(n-3) - f(n-4) } }
Several board test data:
Input 1 1 2 4 9 20 40 90 Output 1 0 10 0 Input 2 2 4 8 16 32 64 128 256 512 2 4 8 16 32 64 128 256 512 Output 2 0 0 0 0 0 0 0 1
Code From:
https://blog.csdn.net/qq_36876305/article/details/80275708
https://blog.csdn.net/running_acmer/article/details/82722111
Data From:
https://www.cnblogs.com/zhouzhendong/p/Berlekamp-Massey.html