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BM-线性递推板子

时间:2019-03-12 09:27:21      阅读:157      评论:0      收藏:0      [点我收藏+]

标签:i++   ack   fir   def   sse   end   names   线性   type   

//杜教BM #include<bits/stdc++.h> using namespace std; #define rep(i,a,n) for (int i=a;i<n;i++) #define per(i,a,n) for (int 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) ((int)(x).size()) typedef vector<int> VI; typedef long long ll; typedef pair<int,int> 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; } ll _,n; namespace linear_seq { const int N=10010; ll res[N],base[N],_c[N],_md[N]; vector<ll> 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) { 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))); } }; ll f[205]; int main() { ll n,m; scanf("%lld%lld",&n,&m); for(int i=1;i<=m;i++) f[i]=1; for(int i=m;i<=200;i++) f[i]=(f[i-1]+f[i-m])%mod; vector<int>v; n++; for(int i=1;i<=200;i++) v.push_back(f[i]); printf("%lld\n",linear_seq::gao(v,n-1)%mod); }

BM-线性递推板子

标签:i++   ack   fir   def   sse   end   names   线性   type   

原文地址:https://blog.51cto.com/14093713/2361531

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