CF 438E The Child and Binary Tree

BZOJ 3625

吐槽

BZOJ上至今没有卡过去，太慢了卡得我不敢交了……

一件很奇怪的事情就是不管是本地还是自己上传数据到OJ测试都远远没有到达时限。

本题做法

设$f_i$表示权值为$i$的二叉树的个数，因为一棵二叉树可以通过左右儿子构建起来转移，我们可以得到转移：

$$f_w = sum_{x, y, w - (x + y) in c} f_x * f_y$$

注意到左右子树可以为空，所以$f_0 = 1$。

很容易发现这是一个卷积的形式，我们尝试把它写得好看一点。

先把物品写成生成函数的形式，记

$$G(x) = sum_{i = 0}^{m}[i in c]x^i$$

题目保证了$G(0) = 0$。

再用$F(x)$表示权值为$x$的二叉树的个数，有

$$F(n) = sum_{i = 1}^{n}G(i)sum_{j = 1}^{n - i}F(j)F(n - i - j)$$

$$F(n) = (G * F ^ 2)(n)$$

发现多项式$F$除了第$0$项每一项都可以由上面这个式子得到，而$F(0) = 1$。

得到

$$F = G * F ^ 2 + 1$$

解一元二次方程，

$$F = frac{2}{1 pm sqrt{1 - 4G}}$$

考虑到$G(0) = 0$，$F(0) = 1$，所以取加号。

剩下就是怎么开根的问题。

多项式开根

假设已经求出了$F_0(x)$使$G(F_0(x)) equiv 0 (mod x^{left lceil frac{n}{2} ight ceil})$成立，要求$F(x)$使$G(F(x)) equiv 0 (mod x^n)$成立。

在多项式$exp$那里已经提过了这里$G(F(x)) = F(x)^2 - A(x)$。

牛顿迭代的式子拿出来，

$$F(x) equiv F_0(x) - frac{G(F_0(x))}{G'(F_0(x))}( mod x^n)$$

$$G'(F(x)) = 2F(x)$$

代进去

$$F(x) equiv frac{F_0(x)^2 + A(x)}{2F_0(x)}(mod x^n)$$

可以递归计算了。

$$T(n) = T(frac{n}{2}) + O(nlogn)$$

时间复杂度还是$O(nlogn)$，常数极大就是了。

Code：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;

const int N = 1 << 18;

int n, m, a[N];
ll f[N], g[N];

namespace Poly {
    const int L = 1 << 18;
    const ll P = 998244353LL;

    int lim, pos[L];
    ll f[L], g[L], h[L], tmp[L];
    
    inline ll fpow(ll x, ll y) {
        ll res = 1;
        for (x %= P; y > 0; y >>= 1) {
            if (y & 1) res = res * x % P;
            x = x * x % P;
        }
        return res;
    }
    
    const ll inv2 = fpow(2, P - 2);
    
    inline void prework(int len) {
        int l = 0;
        for (lim = 1; lim < len; lim <<= 1, ++l);
        for (int i = 0; i < lim; i++)
            pos[i] = (pos[i >> 1] >> 1) | ((i & 1) << (l - 1));
    }
    
    inline void ntt(ll *c, int opt) {
        for (int i = 0; i < lim; i++)
            if (i < pos[i]) swap(c[i], c[pos[i]]);
        for (int i = 1; i < lim; i <<= 1) {
            ll wn = fpow(3, (P - 1) / (i << 1));
            if (opt == -1) wn = fpow(wn, P - 2);
            for (int len = i << 1, j = 0; j < lim; j += len) {
                ll w = 1;
                for (int k = 0; k < i; k++, w = w * wn % P) {
                    ll x = c[j + k], y = w * c[j + k + i] % P;
                    c[j + k] = (x + y) % P, c[j + k + i] = (x - y + P) % P;
                }
            }
        }
        
        if (opt == -1) {
            ll inv = fpow(lim, P - 2);
            for (int i = 0; i < lim; i++) c[i] = c[i] * inv % P;
        }
    }
    
    void inv(ll *a, ll *b, int len) {
        if (len == 1) {
            b[0] = fpow(a[0], P - 2);
            return;
        }
        
        inv(a, b, (len + 1) >> 1);
        prework(len << 1);
        for (int i = 0; i < lim; i++) f[i] = g[i] = 0;
        for (int i = 0; i < len; i++) f[i] = a[i], g[i] = b[i];
        ntt(f, 1), ntt(g, 1);
        for (int i = 0; i < lim; i++)
            g[i] = g[i] * (2LL - g[i] * f[i] % P + P) % P;
        ntt(g, -1);
        
        for (int i = 0; i < len; i++) b[i] = g[i];
    }
    
    inline void direv(ll *c, int len) {
        for (int i = 0; i < len - 1; i++) c[i] = c[i + 1] * (i + 1) % P;
        c[len - 1] = 0;
    }
    
    inline void integ(ll *c, int len) {
        for (int i = len - 1; i > 0; i--) c[i] = c[i - 1] * fpow(i, P - 2) % P;
        c[0] = 0;
    }
    
    inline void ln(ll *a, ll *b, int len) {
        for (int i = 0; i < len; i++) h[i] = 0;
        inv(a, h, len);
                
        for (int i = 0; i < len; i++) b[i] = a[i];
        direv(b, len);    

        prework(len << 1);
        ntt(h, 1), ntt(b, 1);
        for (int i = 0; i < lim; i++) b[i] = b[i] * h[i] % P;
        ntt(b, -1);
        
        integ(b, len);
    }
    
    ll F[L], G[L], H[L];
    void exp(ll *a, ll *b, int len) {
        if (len == 1) {
            b[0] = 1;
            return;
        }
        exp(a, b, (len + 1) >> 1);
        
        ln(b, F, len);
        F[0] = (a[0] % P + 1 - F[0] + P) % P;
        for (int i = 1; i < len; i++) F[i] = (a[i] - F[i] + P) % P;
        
        prework(len << 1);
        for (int i = len; i < lim; i++) F[i] = 0;
        for (int i = 0; i < lim; i++) G[i] = 0;
        for (int i = 0; i < len; i++) G[i] = b[i];
        ntt(F, 1), ntt(G, 1);
        for (int i = 0; i < lim; i++) F[i] = F[i] * G[i] % P;
        ntt(F, -1);
        
        for (int i = 0; i < len; i++) b[i] = F[i];
    }
    
    void sqr(ll *a, ll *b, int len) {
        if (len == 1) {
            b[0] = a[0];
            return;
        }
        
        sqr(a, b, (len + 1) >> 1);
        
        for (int i = 0; i < len; i++) H[i] = 0;
        inv(b, H, len);
        
        prework(len << 1);
        for (int i = 0; i < lim; i++) F[i] = G[i] = 0;
        for (int i = 0; i < len; i++) F[i] = a[i], G[i] = b[i];
        ntt(F, 1), ntt(G, 1), ntt(H, 1);
        for (int i = 0; i < lim; i++) 
            F[i] = (G[i] * G[i] % P + F[i] + P) % P * inv2 % P * H[i] % P;
        ntt(F, -1);
        
        for (int i = 0; i < len; i++) b[i] = F[i];
    }
    
};

using Poly :: fpow;
using Poly :: P;

/*  template <typename T>
inline void read(T &X) {
    X = 0; char ch = 0; T op = 1;
    for (; ch > '9'|| ch < '0'; ch = getchar())
        if (ch == '-') op = -1;
    for (; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}    */

namespace Fread {
    const int L = 1 << 15;
    
    char buffer[L], *S, *T;
    
    inline char Getchar() {
        if(S == T) {
            T = (S = buffer) + fread(buffer, 1, L, stdin);
            if(S == T) return EOF;
        }
        return *S++;
    }
    
    template <class T> 
    inline void read(T &X) {
        char ch; T op = 1;
        for(ch = Getchar(); ch > '9' || ch < '0'; ch = Getchar())
            if(ch == '-') op = -1;
        for(X = 0; ch >= '0' && ch <= '9'; ch = Getchar()) 
            X = (X << 1) + (X << 3) + ch - '0'; 
        X *= op;
    }
    
} using namespace Fread;   

namespace Fwrite {
    const int L = 2e6 + 5;

    int bufp = 0;
    char buf[L]; 
      
    template <typename T>
    inline void write(T x) {
        if(!x) buf[++bufp] = '0';
        
        if(x < 0) {
            buf[++bufp] = '-';
            x = -x;
        }
        int st[15], tp = 0;
        for(; x; x /= 10) st[++tp] = x % 10;
        for(; tp; tp--) buf[++bufp] = (st[tp] + '0');
        buf[++bufp] = '
';
    }   
    

} using namespace Fwrite;

template <typename T>
inline void inc(T &x, T y) {
    x += y;
    if (x >= P) x -= P;
}

template <typename T>
inline void sub(T &x, T y) {
    x -= y;
    if (x < 0) x += P;
}

int main() {
/*    #ifndef ONLINE_JUDGE
        freopen("Sample.txt", "r", stdin);
    #endif    */
    
//    freopen("3625.in", "r", stdin);
//    freopen("3625.out", "w", stdout);
    
    read(n), read(m);
    for (int x, i = 1; i <= n; i++) {
        read(x);
        if (x <= m) g[x] = 1;
    }
    
/*    for (int i = 0; i <= m; i++)
        printf("%lld%c", g[i], " 
"[i == m]);   */
    
    g[0] = 1;
    for (int i = 1; i <= m; i++) g[i] = (P - 4LL * g[i] % P) % P;
    Poly :: sqr(g, f, m + 1);
    
/*    for (int i = 0; i <= m; i++)
        printf("%lld%c", g[i], " 
"[i == m]);   
    
    for (int i = 0; i <= m; i++)
        printf("%lld%c", f[i], " 
"[i == m]);   */

    inc(f[0], 1LL);
    for (int i = 0; i <= m; i++) g[i] = 0;
    Poly :: inv(f, g, m + 1);
    
/*    for (int i = 0; i <= m; i++)
        printf("%lld%c", f[i], " 
"[i == m]);   */

    
    for (int i = 1; i <= m; i++) {
//        write(g[i] * 2LL % P);
//        printf("
");
        printf("%lld
", (g[i] + g[i]) % P);
    }
    
//    fwrite(buf + 1, 1, bufp, stdout);    
    return 0;
}