Solution -「洛谷 P4451」「国家集训队」整数的 lqp 拆分

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;
}