杜教板子（BM) 线性递推式（板子） 杜教板子（BM) 线性递推式

杜教板子（BM) 线性递推式

解决传统线性递推式神器

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cmath>
  4 #include <algorithm>
  5 #include <vector>
  6 #include <string>
  7 #include <map>
  8 #include <set>
  9 #include <cassert>
 10 #include<bits/stdc++.h>
 11 using namespace std;
 12 #define rep(i,a,n) for (int i=a;i<n;i++)
 13 #define per(i,a,n) for (int i=n-1;i>=a;i--)
 14 #define pb push_back
 15 #define mp make_pair
 16 #define all(x) (x).begin(),(x).end()
 17 #define fi first
 18 #define se second
 19 #define SZ(x) ((int)(x).size())
 20 typedef vector<int> VI;
 21 typedef long long ll;
 22 typedef pair<int,int> PII;
 23 const ll mod=1000000007;
 24 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;}
 25 // head
 26 
 27 int _,n;
 28 namespace linear_seq {
 29     const int N=10010;
 30     ll res[N],base[N],_c[N],_md[N];
 31 
 32     vector<int> Md;
 33     void mul(ll *a,ll *b,int k) {
 34         rep(i,0,k+k) _c[i]=0;
 35         rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
 36         for (int i=k+k-1;i>=k;i--) if (_c[i])
 37             rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
 38         rep(i,0,k) a[i]=_c[i];
 39     }
 40     int solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
 41 //        printf("%d\n",SZ(b));
 42         ll ans=0,pnt=0;
 43         int k=SZ(a);
 44         assert(SZ(a)==SZ(b));
 45         rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
 46         Md.clear();
 47         rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
 48         rep(i,0,k) res[i]=base[i]=0;
 49         res[0]=1;
 50         while ((1ll<<pnt)<=n) pnt++;
 51         for (int p=pnt;p>=0;p--) {
 52             mul(res,res,k);
 53             if ((n>>p)&1) {
 54                 for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
 55                 rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
 56             }
 57         }
 58         rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
 59         if (ans<0) ans+=mod;
 60         return ans;
 61     }
 62     VI BM(VI s) {
 63         VI C(1,1),B(1,1);
 64         int L=0,m=1,b=1;
 65         rep(n,0,SZ(s)) {
 66             ll d=0;
 67             rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
 68             if (d==0) ++m;
 69             else if (2*L<=n) {
 70                 VI T=C;
 71                 ll c=mod-d*powmod(b,mod-2)%mod;
 72                 while (SZ(C)<SZ(B)+m) C.pb(0);
 73                 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
 74                 L=n+1-L; B=T; b=d; m=1;
 75             } else {
 76                 ll c=mod-d*powmod(b,mod-2)%mod;
 77                 while (SZ(C)<SZ(B)+m) C.pb(0);
 78                 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
 79                 ++m;
 80             }
 81         }
 82         return C;
 83     }
 84     int gao(VI a,ll n) {
 85         VI c=BM(a);
 86         c.erase(c.begin());
 87         rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;
 88         return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
 89     }
 90 };
 91 
 92 int main() {
 93     while (~scanf("%d",&n)) {
 94         vector<int>v;
 95         v.push_back(1);
 96         v.push_back(2);
 97         v.push_back(4);
 98         v.push_back(7);
 99         v.push_back(13);
100         v.push_back(24);
101         //VI{1,2,4,7,13,24}
102         printf("%d\n",linear_seq::gao(v,n-1));
103     }
104 }
105 
106 BM

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