斐波拉契数列前n项和的求法

一、递推关系构建系数矩阵–矩阵快速幂

{\begin{cases} S [n] = S [n - 1] + f a c [n] \\ f a c [n + 1] = f a c [n] + f a c [n - 1] \\ f a c [n] = f a c [n] \end{cases}

$\begin{cases} S[n] = S[n-1] + fac[n] \\ &&&&\\ fac[n+1] = fac[n] + fac[n-1] \\&&&&\\ fac[n] = fac[n] \end{cases}$

系数矩阵如下：

[\begin{matrix} S [n] \\ f a c [n + 1] \\ f a c [n] \end{matrix}] = [\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{matrix}] * [\begin{matrix} S [n - 1] \\ f a c [n] \\ f a c [n - 1] \end{matrix}]

$\left[ \begin{matrix} S[n] \\ \\ fac[n+1]\\ \\ fac[n] \end{matrix} \right] = \left[ \begin{matrix} 1&1&0\\ \\ 0&1&1\\ \\ 0&1&0 \end{matrix}\right] * \left[ \begin{matrix} S[n-1]\\ \\ fac[n]\\ \\ fac[n-1] \end{matrix} \right]$
递推得到：

[\begin{matrix} S [n] \\ f a c [n + 1] \\ f a c [n] \end{matrix}] = {[\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{matrix}]}^{n} * [\begin{matrix} S [0] \\ f a c [1] \\ f a c [0] \end{matrix}]

$\left[ \begin{matrix} S[n] \\ \\ fac[n+1]\\ \\ fac[n] \end{matrix} \right] = \left[ \begin{matrix} 1&1&0\\ \\ 0&1&1\\ \\ 0&1&0 \end{matrix}\right] ^{n}* \left[ \begin{matrix} S[0]\\ \\ fac[1]\\ \\ fac[0] \end{matrix} \right]$

[\begin{matrix} S [0] \\ f a c [1] \\ f a c [0] \end{matrix}] = [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}]

$\left[ \begin{matrix} S[0]\\ \\ fac[1]\\ \\ fac[0] \end{matrix} \right] = \left[ \begin{matrix} 1\\ \\ 1\\ \\ 1 \end{matrix} \right]$

所以我们需要求出系数矩阵的n次方，S[n] = 系数矩阵第一行的和
牛客OI赛制测试赛3 D

#include <bits/stdc++.h>
#define llt long long
using namespace std;

const llt mod = 998244353;
llt a[3][3];
llt A[3][3]={
               {1,1,0},
               {0,1,1},
               {0,1,0}
            };

llt Ans[3][3]={{1,0,0},{0,1,0},{0,0,1}};
llt ans[3][3];

// 矩阵乘法  b = b*c
void mult(llt b[][3],llt c[][3]){
    llt tmp[3][3];
    for(int i=0;i<3;++i)
      for(int j=0;j<3;++j){
          tmp[i][j] = 0 ;
          for(int k=0;k<3;++k){
             tmp[i][j] += b[i][k]*c[k][j]%mod;
             tmp[i][j] %= mod;
          }
      }
    for(int i=0;i<3;++i)
      for(int j=0;j<3;++j)
        b[i][j] = tmp[i][j];
}
//块速矩阵幂
void quick_mod(llt t){
    for(;t;t>>=1,mult(a,a))
        if(t&1) mult(ans,a);
}

int main(){
    llt n;

    while(~scanf("%lld",&n)){
         for(int i=0;i<3;++i)
           for(int j=0;j<3;++j)
              a[i][j] = A[i][j],ans[i][j]=Ans[i][j];
         quick_mod(n);
         cout<<(ans[0][0]+ans[0][1]+ans[0][2])%mod<<endl;
    }

}

二、线性递推式—杜教BM模板

{\begin{cases} S [n] = S [n - 1] + f a c [n] \\ f a c [n] = f a c [n - 1] + f a c [n - 2] \end{cases}

$\begin{cases} S[n] = S[n-1] + fac[n] \\ &&&&\\ fac[n] = fac[n-1] + fac[n-2] \\ \end{cases}$

得到S[n] 为线性递推式！
输入前0~8项数：1,2,4,7,12,20,33,54,88；(n可以为0）
输入n，即可得到答案！

牛客OI赛制测试赛3 D
前面的1不能删去，因为n可以为0！

#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cassert>
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=998244353;
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<ll> Md;
    void mul(ll *a,ll *b,ll 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;
        ll 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() {

        scanf("%lld",&n);
        printf("%d\n",linear_seq::gao(VI{1,2,4,7,12,20,33,54,88},n));

}

三、斐波拉契n项和公式

$S [n] = f a c [n + 2] - 1$ $S[n] = fac[n+2] -1$

\begin{aligned} (1) & 证 明 ： S [n] & = f a c_{0} + f a c_{1} + f a c_{2} + f a c_{3} + \dots \dots + f a c_{n} \\ (2) & = 1 + f a c_{0} + f a c_{1} + f a c_{2} + f a c_{3} + \dots \dots + f a c_{n} - 1 \\ (3) & = f a c_{1} + f a c_{0} + f a c_{1} + f a c_{2} + f a c_{3} + \dots \dots + f a c_{n} - 1 \\ (4) & = (f a c_{1} + f a c_{0}) + f a c_{1} + f a c_{2} + f a c_{3} + \dots \dots + f a c_{n} - 1 \\ (5) & = (f a c_{2} + f a c_{1}) + f a c_{2} + f a c_{3} + \dots \dots + f a c_{n} - 1 \\ (6) & = (f a c_{3} + f a c_{2}) + f a c_{3} + \dots \dots + f a c_{n} - 1 \\ (7) & = (f a c_{4} + f a c_{3}) + \dots \dots + f a c_{n} - 1 \\ (8) & \dots \dots \\ (9) & = f a c_{n + 1} + f a c_{n} - 1 \\ (10) & = f a c_{n + 2} - 1 \end{aligned}

$\begin{align}证明： S[n] &= fac_0+fac_1+fac_2+fac_3+……+fac_n\tag{1}\\ &= 1+fac_0+fac_1+fac_2+fac_3+……+fac_n-1\tag{2}\\ &= fac_1+fac_0+fac_1+fac_2+fac_3+……+fac_n-1\tag{3}\\ &= (fac_1+fac_0)+fac_1+fac_2+fac_3+……+fac_n-1\tag{4}\\ &= ( fac_2+fac_1)+fac_2+fac_3+……+fac_n-1\tag{5}\\ &= ( fac_3+fac_2)+fac_3+……+fac_n-1\tag{6}\\ &= ( fac_4+fac_3)+……+fac_n-1\tag{7}\\ &……\tag{8}\\ &=fac_{n+1}+fac_n-1\tag{9}\\ &= fac_{n+2}-1\tag{10}\\ \end{align}$
求第n+2项斐波拉契！

{\begin{cases} f a c [n] = f a c [n - 1] + f a c [n - 2] \\ f a c [n - 1] = f a c [n - 1] \end{cases}

$\begin{cases} fac[n] = fac[n-1] + fac[n-2] \\ &&&&\\ fac[n-1] = fac[n-1] \end{cases}$
系数矩阵如下：

[\begin{matrix} f a c [n] \\ f a c [n - 1] \end{matrix}] = [\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}] * [\begin{matrix} f a c [n - 1] \\ f a c [n - 2] \end{matrix}]

$\left[ \begin{matrix} fac[n] \\ \\ fac[n-1]\\ \end{matrix} \right] = \left[ \begin{matrix} 1&1\\ \\ 1&0 \end{matrix}\right] * \left[ \begin{matrix} fac[n-1]\\ \\ fac[n-2] \end{matrix} \right]$
递推得到：

[\begin{matrix} f a c [n] \\ f a c [n - 1] \end{matrix}] = {[\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}]}^{n - 1} * [\begin{matrix} f a c [1] \\ f a c [0] \end{matrix}]

$\left[ \begin{matrix} fac[n] \\ \\ fac[n-1]\\ \end{matrix} \right] = \left[ \begin{matrix} 1&1\\ \\ 1&0 \end{matrix}\right] ^{n-1}* \left[ \begin{matrix} fac[1]\\ \\ fac[0] \end{matrix} \right]$

[\begin{matrix} f a c [1] \\ f a c [0] \end{matrix}] = [\begin{matrix} 1 \\ 1 \end{matrix}]

$\left[ \begin{matrix} fac[1]\\ \\ fac[0] \end{matrix} \right] = \left[ \begin{matrix} 1\\ \\ 1 \end{matrix} \right]$

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即fac[n] = 系数矩阵n-1方的第一行之和！

#include <bits/stdc++.h>
#define llt long long
using namespace std;

const llt mod = 998244353;
llt a[2][2];
llt A[2][2]={
    {1,1},
    {1,0}
};

llt Ans[2][2]={{1,0},{0,1}};
llt ans[2][2];
void mult(llt b[][2],llt c[][2]){
    llt tmp[2][2];
    for(int i=0;i<2;++i)
      for(int j=0;j<2;++j){
          tmp[i][j] = 0 ;
          for(int k=0;k<2;++k){
             tmp[i][j] += b[i][k]*c[k][j]%mod;
             tmp[i][j] %= mod;
          }
      }
    for(int i=0;i<2;++i)
      for(int j=0;j<2;++j)
        b[i][j] = tmp[i][j];
}
void quick_mod(llt t){
    for(;t;t>>=1,mult(a,a))
        if(t&1) mult(ans,a);
}

int main(){
    llt n;

    while(~scanf("%lld",&n)){
         for(int i=0;i<2;++i)
           for(int j=0;j<2;++j)
              a[i][j] = A[i][j],ans[i][j]=Ans[i][j];
         quick_mod(n+1);
         cout<<(ans[0][0]+ans[0][1]-1+mod)%mod<<endl;
    }

}