杜教BM遞推
阿新 • • 發佈:2018-12-08
#include<bits/stdc++.h> using namespace std; #define rep(i,a,n) for (int i=a;i<n;i++) #define pb push_back #define mp make_pair #define SZ(x) ((int)(x).size()) typedef vector<int> VI; typedef long long ll; const ll mod=1000000007; 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;}//快速冪A的B次方 ll n; namespace linear_seq { const int N=10010; ll res[N],base[N],_c[N],_md[N]; vector<int> Md; void mul(ll *a,ll *b,int 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 (int 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]; } int solve(ll n,VI a,VI b) { // a 係數 b 初值 b[n+1]=a[0]*b[n]+... ll ans=0,pnt=0; int 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 (int p=pnt;p>=0;p--) { mul(res,res,k); if ((n>>p)&1) { for (int 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; } } rep(i,0,k) ans=(ans+res[i]*b[i])%mod; if (ans<0) ans+=mod; return ans; } VI BM(VI s) { VI C(1,1),B(1,1); int L=0,m=1,b=1; rep(n,0,SZ(s)) { ll d=0; rep(i,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; } int 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() { //求f(n)=f(n-3)+f(n-1) int a[100],i;vector<int>v;//設記錄A陣列,然後賦給V就可以求結果了 a[1] = a[2] = a[3] = 1;for(i = 4;i <= 99;i ++)a[i] = (a[i-3]+a[i-1])%mod; for(i = 1;i <= 99;i ++) v.push_back(a[i]); int nCase;scanf("%d", &nCase); while(nCase--){ scanf("%lld", &n);//求出數列第N項 printf("%lld\n",1LL * linear_seq::gao(v,n-1) % mod);//傳進V向量,0編號~N-1 } return 0; }