51Nod 1242 斐波那契数列的第N项 ——————杜教筛，BM板子

不念不忘少年蓝@ 2022-05-09 10:18 125阅读 0赞

# 1242 斐波那契数列的第N项 #

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斐波那契数列的定义如下：

F ( 0 ) = 0 F ( 1 ) = 1 F ( n ) = F ( n − 1 ) + F ( n − 2 ) ( n > = 2 ) F(0) = 0 \\\\ F(1) = 1\\\\ F(n) = F(n - 1) + F(n - 2)\\qquad (n >= 2) F(0)=0F(1)=1F(n)=F(n−1)\+F(n−2)(n>=2)

( 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , 233 , 377 , . . . ) (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ...) (1,1,2,3,5,8,13,21,34,55,89,144,233,377,...)  
给出 n n n，求 F ( n ) F(n) F(n)，由于结果很大，输出 &ThickSpace; F ( n ) % 1000000009 &ThickSpace;&ThickSpace; \\;F(n) \\% 1000000009\\;\\; F(n)%1000000009的结果即可。

**Input**  
输入1个数n(1 <= n <= 10^18)。  
**Output**  
输出 F ( n ) % 1000000009 F(n) \\% 1000000009 F(n)%1000000009的结果。

**Input**示例  
11  
**Output**示例  
89

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BM线性筛出递推式的第n项

#include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <vector>
    #include <string>
    #include <map>
    #include <set>
    #include <cassert>
    #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=1000000009;
    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
    
    int _;
    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 系数 b 初值 b[n+1]=a[0]*b[n]+...
    //        printf("%d\n",SZ(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)));
        }
    };
    
    int main() {
        
        while (~scanf("%lld",&n)) {
        
            vector<int>v;
            v.push_back(0);
            v.push_back(1);
            v.push_back(1);
            v.push_back(2);
            v.push_back(3);
            v.push_back(5);
            //VI{1,2,4,7,13,24}
    
            printf("%d\n",linear_seq::gao(v,n));
        }
    }