【LOJ #6268】分拆數（DP / 生成函數）

可以參見$EI$的博客

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傳送門

首先有一個$\exp$的$O(nlogn)$做法
這裏

考慮$Ferrers圖$
從左上角向右下截一個最大的正方形
設正方形邊長爲$h$
那麼剩下兩部分就都是$\le h$的整數拆分

即可以列出分拆數的生成函數$\prod_{i\geq 1}\frac{1}{1-x^i}=\sum_{h\geq 1}x^{h^2}(\prod_{k=1}^h\frac 1{1-x^k})$

由於有$h^2$所以只用考慮$h\le \sqrt n$時候計算即可
複雜度$O(n\sqrt n)$
吊打$\exp$

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define ll long long
#define fi first
#define se second
#define bg begin
cs int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ib==ob)?EOF:*ib++;
}
inline int read(){
    char ch=gc();
    int res=0;bool 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;
}
inline ll readll(){
    char ch=gc();
    ll res=0;bool 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;
}
inline int readstring(char *s){
	int top=0;char ch=gc();
	while(isspace(ch))ch=gc();
	while(!isspace(ch)&&ch!=EOF)s[++top]=ch,ch=gc();
	return top;
}
template<typename tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<typename tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int mod=998244353;
inline int add(int a,int b){a+=b-mod;return a+(a>>31&mod);}
inline int dec(int a,int b){return a-b<0?a-b+mod:a-b;}
inline int mul(int a,int b){static ll r;r=(ll)a*b;return (r>=mod)?(r%mod):r;}
inline void Add(int &a,int b){a+=b-mod;a+=a>>31&mod;}
inline void Dec(int &a,int b){a-=b;a+=a>>31&mod;}
inline void Mul(int &a,int b){static ll r;r=(ll)a*b;a=(r>=mod)?(r%mod):r;}
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);}
cs int N=100005;
int f[N],g[N];
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	int n=read();
	int b=sqrt(n);
	f[0]=g[0]=1;
	for(int i=1,p;i<=b;i++){
		p=i*i;
		for(int tt=1;tt<=2;tt++)
		for(int j=i;j+p<=n;j++)
		Add(f[j],f[j-i]);
		for(int j=p;j<=n;j++)
		Add(g[j],f[j-p]);
	}
	for(int i=1;i<=n;i++)cout<<g[i]<<"\n";
	return 0;
}