CF755GPolandBall and Many Other Balls

发表于 | 阅读数

有一排$n(n\le 10^9)$ 个球,定义一个组可以只包含一个球或者包含两个相邻的球.现在一个球只能分到一个组中,求从这些球中取出 $k(k<2^{15})$组的方案数.

设$dp[i][j]$,为把前$i$个球分成$j$组的方案数。

显然

把$dp[i]$看出一个多项式，即$dp[i]=(x+1)dp[i-1]+xdp[i-2]$

倍增FFT

递推式已经得出来。考虑倍增式。
显然

  • 两堆不影响$dp[i+j][k]=\sum_{a+b=k}dp[i][b]*dp[j][a]$
  • 两堆中有$1+1$,$dp[i+j][k+1]=\sum_{a+b=k}dp[i-1][b]*dp[j-1][a]$

维护三个$(dp[x-2],dp[x-1],dp[x])\rightarrow(dp[x-1+x-1],dp[x-1+x],dp[x+x])$
类似快速幂的方法(遇到$1$往末尾$+1$,即递推，都有$<<1$)

特征方程

显然把$x$看成一阶特征方程的递推式，即

  • 注意由于常数项为$0$，可以进行开根，快速幂（简单版）
代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 1e6 + 10;

const int P = 998244353, gi = 3;
const double pi = acos(-1.0);
int inc(int x, int y, int mo)
{
    if (x + y >= mo)
        return (x + y - mo);
    else
        return (x + y);
}

int del(int x, int y, int mo)
{
    if (x - y < 0)
        return (x - y + mo);
    else
        return (x - y);
}
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}

int qpow(int a, int x, int mo)
{
    int res = 1;
    while (x)
    {
        if (x & 1)
            res = 1ll * res * a % mo;
        x >>= 1;
        a = 1ll * a * a % mo;
    }
    return res;
}
int rev[N];
void NTT(int *A, int n, int inv)
{
    for (int i = 0; i < n; i++)
        if (i < rev[i])
            swap(A[i], A[rev[i]]);
    for (int l = 1; l < n; l <<= 1)
    {
        int tt = qpow(gi, (P - 1) / (l << 1), P);
        int temp = (inv == 1 ? tt : qpow(tt, P - 2, P));
        for (int i = 0; i < n; i += (l << 1))
        {
            int omega = 1;
            for (int j = 0; j < l; j++, omega = 1ll * omega * temp % P)
            {
                int x = A[i + j], y = 1ll * omega * A[i + j + l] % P;
                A[i + j] = inc(x, y, P);
                A[i + j + l] = del(x, y, P);
            }
        }
    }
    int invv = qpow(n, P - 2, P);
    if (inv == -1)
        for (int i = 0; i < n; i++)
            A[i] = 1ll * A[i] * invv % P;
}
void NTTX(int *a, int n, int *b, int m)
{
    int ML = 1, bit = 0;
    while (ML < n + m)
        ML <<= 1, bit++;
    for (int i = 0; i < ML; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    NTT(a, ML, 1);
    NTT(b, ML, 1);
    for (int i = 0; i < ML; i++)
        a[i] = 1ll * a[i] * b[i] % P;
    NTT(a, ML, -1);
}
int Inv2;
int C[N], D[N];

void Finv(int *a, int *b, int n)
{
    b[0] = qpow(a[0], P - 2, P);
    int len, ML;
    int bit = 0;

    for (len = 1; len < (n << 1); len <<= 1)
    {
        ML = len << 1;
        bit++;
        for (int i = 0; i < len; i++)
            C[i] = a[i], D[i] = b[i];
        for (int i = 0; i < ML; i++)
            rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
        NTT(C, ML, 1), NTT(D, ML, 1);
        for (int i = 0; i < ML; i++)
            b[i] = ((2ll - 1ll * C[i] * D[i] % P) * D[i] % P + P) % P;
        NTT(b, ML, -1);
        for (int i = len; i < ML; i++)
            b[i] = 0;
    }
    for (int i = 0; i < len; i++)
        C[i] = D[i] = 0;
    for (int i = n; i < len; i++)
        b[i] = 0;
}
int E[N], F[N];

void Sqrt(int *a, int *b, int n)
{
    Inv2 = qpow(2, P - 2, P);
    b[0] = 1;
    int bit = 0;
    int len;

    for (len = 1; len < (n << 1); len <<= 1)
    {
        int ML = len << 1;
        bit++;
        for (int i = 0; i < len; i++)
            E[i] = a[i];
        Finv(b, F, len);
        for (int i = 0; i < ML; i++)
            rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
        NTT(E, ML, 1), NTT(F, ML, 1);
        for (int i = 0; i < ML; i++)
            E[i] = 1ll * E[i] * F[i] % P;
        NTT(E, ML, -1);
        for (int i = 0; i < len; i++)
            b[i] = 1LL * (b[i] + E[i]) % P * Inv2 % P;
        for (int i = len; i < ML; i++)
            b[i] = 0;
    }
    for (int i = 0; i < len; i++)
        E[i] = F[i] = 0;
    for (int i = n; i < len; i++)
        b[i] = 0;
}

int aa[N], ia[N]; // aa 表示a的导数，ia表示a的逆
void Ln(int *a, int *res, int n)
{
    for (int i = 0; i < n << 1; i++)
        aa[i] = ia[i] = 0;
    for (int i = 1; i < n; i++)
        aa[i - 1] = 1ll * i * a[i] % P;
    Finv(a, ia, n);
    NTTX(aa, n, ia, n);
    res[1] = 0;
    for (int i = 1; i < n; i++)
        res[i] = 1ll * aa[i - 1] * qpow(i, P - 2, P) % P;
}
int G[N], H[N], M[N];
void EXP(int *a, int *b, int n)
{
    Inv2 = qpow(2, P - 2, P);
    b[0] = 1;
    int bit = 0;
    int len;

    for (len = 1; len < (n << 1); len <<= 1)
    {
        int ML = len << 1;
        bit++;
        for (int i = 0; i < len << 1; i++)
            H[i] = G[i] = M[i] = 0;
        for (int i = 0; i < len; i++)
            H[i] = a[i];
        for (int i = 0; i < len; i++)
            G[i] = b[i];
        Ln(G, M, len); // M(x)=lin(G)
        for (int i = 0; i < ML; i++)
            rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
        NTT(G, ML, 1), NTT(H, ML, 1), NTT(M, ML, 1);
        for (int i = 0; i < ML; i++)
            b[i] = 1LL * G[i] * (1ll - M[i] + H[i] + P) % P;
        NTT(b, ML, -1);
        for (int i = len; i < ML; i++)
            b[i] = 0;
    }
    for (int i = 0; i < len; i++)
        G[i] = H[i] = M[i] = 0;
    for (int i = n; i < len; i++)
        b[i] = 0;
}
int c[N];
void Fpow(int *f, int k, int n, int *g)
{
    Ln(f, c, n);
    for (int i = 0; i < n; i++)
        c[i] = 1ll * c[i] * k % P;
    EXP(c, g, n);
}
int f[N], g[N], ff[N], gg[N];
int main()
{
    int n = read(), m = read();
    int k = min(n, m);
    f[0] = 1, f[1] = 6, f[2] = 1;
    g[0] = 1, g[1] = 6, g[2] = 1;
    Sqrt(f, ff, k + 1);
    Sqrt(g, gg, k + 1);
    memset(g, 0, sizeof(g));
    memset(f, 0, sizeof(f));
    Finv(gg, g, k + 1);
    ff[0]++, ff[1]++;
    int inv2 = qpow(2, P - 2, P);
    for (int i = 0; i <= k + 1; i++)
        ff[i] = 1ll * ff[i] % P;
    Fpow(ff, n + 1, k + 1, f);
    NTTX(f, k + 1, g, k + 1);
    for (int i = 1; i <= k; ++i)
        printf("%d ", f[i]);
    for (int i = k + 1; i <= m; ++i)
        printf("0 ");
}