【洛谷P4451】整数的lqp拆分（生成函数）

传送门

考虑设$g(x)$为$ans$的生成函数
那么有
$g(x)=f(x)g(x)+1$
$g(x)=frac{1}{1-f(x)}$
由于$f(x)=f(x)x+f(x)x^2+x$
所以$f(x)=frac{x}{1-x-x^2}$
$g(x)=1+frac{x}{1-2x-x^2}$
那么有$g(x)=2xg(x)+x^2g(x)+x$
于是$ans[i]=2ans[i-1]+ans[i-2]$

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
    char ch=gc();
    int res=0,f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
#define ll long long
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int mod=1e9+7;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?a-=mod:0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?a+=mod:0;}
inline int mul(int a,int b){return 1ll*a*b%mod;}
inline void Mul(int &a,int b){a=1ll*a*b%mod;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
int f[1000005],n;
int main(){
	n=read();
	f[1]=1;
	for(int i=2;i<=n;i++)f[i]=add(add(f[i-1],f[i-1]),f[i-2]);
	cout<<f[n];
}