Description

Link.

求

$$ \sum\prod_{i=1}^{m}F_{a_{i}},(m>0,a_{1},\cdots a_{m}>0,\sum a_{i}=n) $$

Solution

这是一篇不用 $\mathbf{OGF}$ 的题解。

设 $f_{i}$ 为 $i$ 的 $\operatorname{lqp}$ 拆分值。

然后有显然的过不了递推式：

$$ f_{n}=\begin{cases} 1,n=0 \\ \displaystyle \sum_{i=0}^{n-1}F_{n-i}\times f_{i},n\neq0 \end{cases} $$

然后传统艺能错位相减操作一下：

$$ \begin{aligned} f_{n}&=\sum_{i=0}^{n-1}F_{n-i}\times f_{i} \\ f_{n-1}&=\sum_{i=0}^{n-2}F_{n-i-1}\times f_{i} \\ f_{n-2}&=\sum_{i=0}^{n-3}F_{n-i-2}\times f_{i} \end{aligned} \Longrightarrow \begin{aligned} f_{n}-f_{n-1}-f_{n-2}&=\sum_{i=0}^{n-1}F_{n-i}\times f_{i}-\sum_{i=0}^{n-2}F_{n-i-1}\times f_{i}-\sum_{i=0}^{n-3}F_{n-i-2}\times f_{i} \\ f_{n}-f_{n-1}-f_{n-2}&=(F_{2}-F_{1})\times f_{n-2}+F_{1}\times f_{n-1} \end{aligned} \\ \downarrow \\ f_{n}=2f_{n-1}+f_{n-2} $$

递推公式有了，然后矩阵快速幂：

$$ \begin{bmatrix} f_{n} \\ f_{n-1} \end{bmatrix} =\begin{bmatrix} 2f_{n-1}+f_{n-2} \\ f_{n-1} \end{bmatrix} =\begin{bmatrix} f_{n-1} \\ f_{n-2} \end{bmatrix} \times\begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix} $$

这样就可以做了（吗？）：

（code？）

#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
#define mod ( 1000000007 )

using namespace std;
typedef long long LL;

template<typename _T, typename _P>
_T qkpow( _T bas, _T one, _P fur ){
    _T res = one;
    while( fur != 0 ){
        if( fur % 2 == ( _P )1 )    res = bas * res;
        bas = bas * bas;
        fur /= 2;
    }
    return res;
}

template<typename _T>
_T add( _T x, _T y ){ if( y >= mod )    y %= mod; x += y; if( x >= mod )    x -= mod; return x; }

struct bigInt : vector<int>{
    bigInt &check( ){
        while( ! empty( ) && ! back( ) ) pop_back( );
        if( empty( ) )    return *this;
        for( unsigned i = 1; i < size( ); ++ i ){ ( *this )[i] += ( *this )[i - 1] / 10; ( *this )[i - 1] %= 10; }
        while( back( ) >= 10 ){ push_back( back( ) / 10 ); ( *this )[size( ) - 2] %= 10; }
        return *this;
    }
    bigInt( int tpN = 0 ){ push_back( tpN ); check( ); }
};
istream &operator >> ( istream &is, bigInt &tpN ){
    string s;
    is >> s; tpN.clear( );
    for( int i = s.size( ) - 1; i >= 0; --i ) tpN.push_back( s[i] - '0' );
    return is;
}
ostream &operator << ( ostream &os, const bigInt &tpN ){
    if( tpN.empty( ) )    os << 0;
    for( int i = tpN.size( ) - 1; i >= 0; --i )    os << tpN[i];
    return os;
}
bool operator != ( const bigInt &one, const bigInt &another ){
    if( one.size( ) != another.size( ) )    return 1;
    for( int i = one.size( ) - 1; i >= 0; --i ){
        if( one[i] != another[i] )    return 1;
    }
    return 0;
}
bool operator == ( const bigInt &one, const bigInt &another ){
    return ! ( one != another );
}
bool operator < ( const bigInt &one, const bigInt &another ){
    if( one.size( ) != another.size( ) )    return one.size( ) < another.size( );
    for( int i = one.size( ) - 1; i >= 0; --i ){
        if( one[i] != another[i] )    return one[i] < another[i];
    }
    return 0;
}
bool operator > ( const bigInt &one, const bigInt &another ){ return another < one; }
bool operator <= ( const bigInt &one, const bigInt &another ){ return ! (one > another ); }
bool operator >= ( const bigInt &one, const bigInt &another ){ return ! (one < another ); }
bigInt &operator += ( bigInt &one, const bigInt &another ){
    if( one.size( ) < another.size( ) )    one.resize(another.size( ) );
    for( unsigned i = 0; i != another.size( ); ++ i ) one[i] += another[i];
    return one.check( );
}
bigInt operator + ( bigInt one, const bigInt &another ){ return one += another; }
bigInt &operator -= ( bigInt &one, bigInt another ){
    if( one < another )    swap( one, another );
    for( unsigned i = 0; i != another.size( ); one[i] -= another[i], ++ i ){
        if( one[i] < another[i] ){
            unsigned j = i + 1;
            while( ! one[j] ) ++ j;
            while( j > i ){ -- one[j]; one[--j] += 10; }
        }
    }
    return one.check( );
}
bigInt operator - ( bigInt one, const bigInt &another ){ return one -= another; }
bigInt operator * ( const bigInt &one, const bigInt &another ){
    bigInt tpN;
    tpN.assign( one.size( ) + another.size( ) - 1, 0 );
    for( unsigned i = 0; i != one.size( ); ++ i ){
        for( unsigned j = 0; j != another.size( ); ++ j ) tpN[i + j] += one[i] * another[j];
    }
    return tpN.check( );
}
bigInt &operator *= ( bigInt &one, const bigInt &another ){ return one = one * another; }
bigInt divMod( bigInt &one, const bigInt &another ){
    bigInt ans;
    for( int t = one.size( ) - another.size( ); one >= another; -- t ){
        bigInt tpS;
        tpS.assign( t + 1, 0 );
        tpS.back( ) = 1;
        bigInt tpM = another * tpS;
        while( one >= tpM ){ one -= tpM; ans += tpS; }
    }
    return ans;
}
bigInt operator / ( bigInt one, const bigInt &another ){ return divMod(one, another ); }
bigInt &operator /= ( bigInt &one, const bigInt &another ){ return one = one / another; }
bigInt &operator %= ( bigInt &one, const bigInt &another ){ divMod( one, another ); return one; }
bigInt operator % ( bigInt one, const bigInt &another ){ return one %= another; }

struct matrixS{
    int mat[2][2];
    matrixS( int x = 0 ){ memset( mat, x, sizeof( mat ) ); }
    matrixS operator * ( const matrixS &another ) const{
        matrixS res;
        for( int i = 0; i < 2; ++ i ){
            for( int j = 0; j < 2; ++ j ){
                for( int k = 0; k < 2; ++ k )    res.mat[i][j] = add( ( LL )res.mat[i][j], ( LL )mat[i][k] * another.mat[k][j] );
            }
        }
        return res;
    }
} unit, erng;

bigInt N;

void progressBaseInformation( ){
    int unitS[2][2] = { { 1, 0 }, { 0, 1 } };
    memcpy( unit.mat, unitS, sizeof( unitS ) );
    int erngS[2][2] = { { 2, 1 }, { 1, 0 } };
    memcpy( erng.mat, erngS, sizeof( erngS ) );
}

signed main( ){
    ios::sync_with_stdio( 0 ); cin.tie( 0 ); cout.tie( 0 );
    progressBaseInformation( );
    cin >> N; cout << qkpow( erng, unit, N ).mat[1][0] << '\n';
    return 0;
}

不，凉心出题人友好地卡了高精的常数，于是你打开题解，发现 $f_{n}=f_{n\bmod (10^{9}+6)}$，于是你又行了。

$\mathcal{Code}$

#include <cstdio>
#include <cstring>
#include <queue>
#define mod ( 1000000007 )

using namespace std;
typedef long long LL;

template<typename _T>
void read( _T &x ){
    x = 0; char c = getchar( ); _T f = 1;
    while( c < '0' || c > '9' ){ if( c == '-' )    f = -1; c = getchar( ); }
    while( c >= '0' && c <= '9' ){ x = ( ( x << 3 ) + ( x << 1 ) + ( c & 15 ) ) % ( mod - 1 ); c = getchar( ); }
    x *= f;
}

template<typename _T>
void write( _T x ){
    if( x < 0 ){ putchar( '-' ); x = -x; }
    if( x > 9 )    write( x / 10 );
    putchar( x % 10 + '0' );
}

template<typename _T, typename _P>
_T qkpow( _T bas, _T one, _P fur ){
    _T res = one;
    while( fur != 0 ){
        if( fur % 2 == ( _P )1 )    res = bas * res;
        bas = bas * bas;
        fur /= 2;
    }
    return res;
}

template<typename _T>
_T add( _T x, _T y ){ if( y >= mod )    y %= mod; x += y; if( x >= mod )    x -= mod; return x; }

struct matrixS{
    int mat[2][2];
    matrixS( int x = 0 ){ memset( mat, x, sizeof( mat ) ); }
    matrixS operator * ( const matrixS &another ) const{
        matrixS res;
        for( int i = 0; i < 2; ++ i ){
            for( int j = 0; j < 2; ++ j ){
                for( int k = 0; k < 2; ++ k )    res.mat[i][j] = add( ( LL )res.mat[i][j], ( LL )mat[i][k] * another.mat[k][j] );
            }
        }
        return res;
    }
} unit, erng;

LL N;

void progressBaseInformation( ){
    int unitS[2][2] = { { 1, 0 }, { 0, 1 } };
    memcpy( unit.mat, unitS, sizeof( unitS ) );
    int erngS[2][2] = { { 2, 1 }, { 1, 0 } };
    memcpy( erng.mat, erngS, sizeof( erngS ) );
}

signed main( ){
    progressBaseInformation( );
    read( N ); write( qkpow( erng, unit, N ).mat[1][0] ), putchar( '\n' );
    return 0;
}