鬼魅 分拆数

https://loj.ac/problem/6268  模板题。  

一： 生成函数？ 多项式exp？（最慢？） https://www.cnblogs.com/Kong-Ruo/p/10716004.html  （原文）

#include <bits/stdc++.h>
#define LL long long
#define rep(i, s, t) for (register int i = (s), i##end = (t); i <= i##end; ++i)
#define dwn(i, s, t) for (register int i = (s), i##end = (t); i >= i##end; --i)
using namespace std;
inline int read() {
    int x = 0, f = 1;
    char ch;
    for (ch = getchar(); !isdigit(ch); ch = getchar())
        if (ch == '-')
            f = -f;
    for (; isdigit(ch); ch = getchar()) x = 10 * x + ch - '0';
    return x * f;
}
const int maxn = 600010, mod = 998244353;
int F[maxn], G[maxn];
inline int skr(int x, int t) {
    int res = 1;
    for (; t; x = 1LL * x * x % mod, t >>= 1)
        if (t & 1)
            res = 1LL * res * x % mod;
    return res;
}
int r[maxn], lg[maxn], temp[maxn];
int inv[maxn], ifac[maxn], fac[maxn];
inline int skr(int x, LL t) {
    int res = 1;
    while (t) {
        if (t & 1)
            res = 1LL * res * x % mod;
        x = 1LL * x * x % mod;
        t = t >> 1;
    }
    return res;
}
inline void fft(int *a, int n, int type) {
    for (int i = 0; i < n; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (lg[n] - 1));
    for (int i = 0; i < n; i++)
        if (i < r[i])
            swap(a[i], a[r[i]]);
    for (int i = 1; i < n; i <<= 1) {
        int wn = skr(3, (mod - 1) / (i << 1));
        if (type == -1)
            wn = skr(wn, mod - 2);
        // cout << wn << endl;
        for (int j = 0; j < n; j += (i << 1)) {
            int w = 1;
            for (int k = 0; k < i; k++, w = (1LL * (LL)w * (LL)wn) % mod) {
                int x = a[j + k], y = (1LL * (LL)w * (LL)a[j + k + i]) % mod;
                a[j + k] = (x + y) % mod;
                a[j + k + i] = (((x - y) % mod) + mod) % mod;
            }
        }
    }
    if (type == -1) {
        int inv = skr(n, mod - 2);
        for (int i = 0; i < n; i++) a[i] = ((LL)a[i] * (LL)inv) % mod;
    }
}
inline void Inverse(int *a, int *b, int n) {
    if (n == 1) {
        b[0] = skr(a[0], mod - 2);
        return;
    }
    Inverse(a, b, n >> 1);
    memcpy(temp, a, n * sizeof(int));
    memset(temp + n, 0, n * sizeof(int));
    fft(temp, n << 1, 1);
    fft(b, n << 1, 1);
    for (int i = 0; i < (n << 1); i++)
        b[i] = 1LL * b[i] * ((2LL - 1LL * temp[i] * b[i] % mod + mod) % mod) % mod;
    fft(b, n << 1, -1);
    memset(b + n, 0, n * sizeof(int));
}
int c[maxn], d[maxn];
inline void Ln(int *a, int *b, int n) {
    Inverse(a, c, n);
    for (int i = 0; i < n - 1; ++i) d[i] = (LL)(i + 1) * a[i + 1] % mod;
    d[n - 1] = 0;
    fft(c, n << 1, 1);
    fft(d, n << 1, 1);
    for (int i = 0; i < (n << 1); ++i) c[i] = 1LL * d[i] * c[i] % mod;
    fft(c, (n << 1), -1);
    for (int i = 1; i < (n << 1); ++i) b[i] = 1LL * inv[i] * c[i - 1] % mod;
    b[0] = 0;
    for (int i = 0; i < (n << 1); ++i) c[i] = d[i] = 0;
}
int temp_w[maxn], temp_Ln[maxn];
inline void Exp(int *a, int *b, int n) {
    if (n == 1) {
        b[0] = 1;
        return;
    }
    Exp(a, b, n >> 1);
    memcpy(temp_w, b, sizeof(int) * n);
    memset(temp_w + n, 0, sizeof(int) * n);
    Ln(b, temp_Ln, n);
    for (int i = 0; i < n; i++) temp_Ln[i] = (mod + a[i] - temp_Ln[i]) % mod;
    (temp_Ln[0] += 1) %= mod;
    fft(temp_w, n << 1, 1);
    fft(temp_Ln, n << 1, 1);
    for (int i = 0; i < (n << 1); i++) temp_w[i] = 1LL * temp_w[i] * temp_Ln[i] % mod;
    fft(temp_w, n << 1, -1);
    memcpy(b, temp_w, n * sizeof(int));
    memset(b + n, 0, n * sizeof(int));
    memset(temp_w, 0, sizeof(int) * (n << 1));
    memset(temp_Ln, 0, sizeof(int) * (n << 1));
}
int main() {
    int n = read(); lg[0] = -1;
    rep(i, 1, 600000) lg[i] = lg[i >> 1] + 1;
    int len = 1; for(; len <= n; len <<= 1);
    inv[1] = ifac[0] = fac[0] = 1;
    rep(i, 1, len) {
        if (i != 1) inv[i] = -(LL)mod / i * inv[mod % i] % mod;
        inv[i] = ((inv[i] % mod) + mod) % mod;
        ifac[i] = (LL)ifac[i - 1] * inv[i] % mod;
        fac[i] = (LL)fac[i - 1] * i % mod;
    }
    rep(i, 1, n) rep(j, 1, n / i) F[i * j] = (F[i * j] + inv[j]) % mod;
    //rep(i, 1, n) cout << F[i] << " ";
    //cout << endl;
    Exp(F, G, len); rep(i, 1, n) cout << G[i] << '
';
}

二： dp？？ 背包？  （较慢?）  原文：https://www.cnblogs.com/JYYHH/p/8954965.html

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=100005,root=3,ha=998244353,inv=ha/3+1,Base=333;
inline int add(int x,int y){ x+=y; return x>=ha?x-ha:x;}
inline void ADD(int &x,int y){ x+=y; if(x>=ha) x-=ha;}
void W(int x){ if(x>=10) W(x/10); putchar(x%10+'0');}
int A[maxn*4],F[Base+5][maxn],B[maxn*4],r[maxn*4],n,l,S,N,INV;
inline int ksm(int x,int y){ int an=1; for(;y;y>>=1,x=x*(ll)x%ha) if(y&1) an=an*(ll)x%ha; return an;}
 
inline void dp(){
    A[0]=1;
    for(int i=1;i<Base;i++)
        for(int j=i;j<=n;j++) ADD(A[j],A[j-i]);
         
    F[0][0]=1,S=maxn/Base+1,F[1][Base]=1;
    for(int i=1;i<S;i++)
        for(int j=0;j<=n;j++) if(F[i][j]){
            if(j+i<=n) ADD(F[i][j+i],F[i][j]);
            if(j+Base<=n) ADD(F[i+1][j+Base],F[i][j]);
            ADD(B[j],F[i][j]);
        }
    ADD(B[0],1);
}
 
inline void NTT(int *c,int f){
    for(int i=0;i<N;i++) if(i<r[i]) swap(c[i],c[r[i]]);
     
    for(int i=1;i<N;i<<=1){
        int omega=ksm((f==1?root:inv),(ha-1)/(i<<1));
        for(int P=i<<1,j=0;j<N;j+=P){
            int now=1;
            for(int k=0;k<i;k++,now=now*(ll)omega%ha){
                int x=c[j+k],y=c[j+k+i]*(ll)now%ha;
                c[j+k]=add(x,y);
                c[j+k+i]=add(x,ha-y);
            }
        }
    }
     
    if(f==-1) for(int i=0;i<N;i++) c[i]=c[i]*(ll)INV%ha;
}
 
int main(){
    scanf("%d",&n),dp();
     
    for(N=1;N<=(n<<1);N<<=1) l++;
    for(int i=0;i<N;i++) r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
    NTT(A,1),NTT(B,1);
    for(int i=0;i<N;i++) A[i]=A[i]*(ll)B[i]%ha;
    INV=ksm(N,ha-2),NTT(A,-1);
     
    for(int i=1;i<=n;i++) W(A[i]),puts("");
    return 0;
}

三：数论与五边形数定理？ （最快?） 原文：https://blog.csdn.net/weixin_41698125/article/details/79334581

步骤：

    一、构造母函数（像上面那样）并化简成1/[(1-x)*(1-x^2)*(1-x^3)*...]（化简可参见无穷级数）

    二、分析(1-x)*(1-x^2)*(1-x^3)*...（这个函数也叫欧拉函数）的系数。（结论：系数的通项为 k*(3k±1）/2）（这个东西证明起来十分复杂，自行百度五边形数定理）

    三、p（n）与欧拉函数的关系，通过分析n次项系数可得p(n)-p(n-1)-p(n-2)+p(n-5)+p(n-7)-p(n-12)-p(n-15)+…-…=0

    于是我们可以用前几项的p（n）来得到新的p（n），这是效率相当高的方法。时间复杂度o(n^3/2)。

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define LL long long
#define ULL unsigned long long
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define dep(i,j,k) for(int i=k;i>=j;i--)
#define INF 0x3f3f3f3f
#define mem(i,j) memset(i,j,sizeof(i))
#define make(i,j) make_pair(i,j)
#define pb push_back
using namespace std;
const int N = 1e5 + 5;
const LL mod = 998244353;
LL p[N];
LL get(LL n) {
    LL res = 0, k = 1, a = 2, b = 1, s = 1;
    while (n >= a){
        res += s * (p[n - a] % mod + p[n - b] % mod) % mod;
        res = ( res + mod) % mod;
        a += 3 * k + 2;
        b += 3 * k + 1;
        s *= -1;
        k += 1;
    }
    res += (n >= b) ? s * p[n - b] : 0;
    return ( res + mod ) % mod;
}
int main() {
    p[0] = p[1] = 1LL;
    rep(i, 2, N - 5) {
        p[i] = get(1LL * i) % mod;
    }
    int n;
    scanf("%d", &n);
    rep(i, 1, n) {
        printf("%lld
", p[i]);
        //if(p[i] < 0 ) break;
    }
    return 0;
}