# Andrew Stankevich Contest 46 Problem D. Dichromatic Trees

G[i][j]表示根节点为黑色，包含i个节点，black height为j的红黑树个数

$\rm F[i][j]=\sum\limits_{k=0}^{i-1}G[k][j]*G[i-k][j]$

$\rm G[i][j]=\sum\limits_{k=0}^{i-1}(F[k][j-1]+G[k][j-1])*(F[i-k][j-1]+G[i-k][j-1])$

#include <bits/stdc++.h>
#define int long long
using namespace std;

#define rep(i,x,y) for(int i=x;i<=y;i++)

int rd() {
int ret = 0, f = 1;char c;
while (c = getchar(), !isdigit(c))f = c == '-' ? -1 : 1;
while (isdigit(c))ret = ret * 10 + c - '0', c = getchar();
return ret * f;
}

typedef long long ll;
typedef long double ld;

const int inf = 1 << 30;
const int MAXN = 524289+2;
int MOD = 258280327;
const int BASE = 1 << 15;
const ld Pi = acos(-1.0);

struct CP {
ld x, y;
CP(ld xx = 0, ld yy = 0) {
x = xx;
y = yy;
}
} P1[MAXN << 2], P2[MAXN << 2], Q[MAXN << 2];

CP operator+(CP a, CP b) {
return {a.x + b.x, a.y + b.y};
}

CP operator-(CP a, CP b) {
return {a.x - b.x, a.y - b.y};
}

CP operator*(CP a, CP b) {
return {a.x *b.x - a.y * b.y, a.x *b.y + a.y * b.x};
}

int limit, r[MAXN << 2];

ll qpow(ll x, ll y) {
ll ret = 1, base = x;
while (y) {
if (y & 1)
ret = ret * base % MOD;
base = base * base % MOD;
y >>= 1;
}
return ret;
}

void FFT(CP *A, int type) {
for (int i = 0; i < limit; i++)
if (i < r[i])
swap(A[i], A[r[i]]);
for (int mid = 1; mid < limit; mid <<= 1) {
CP Wn(cos(Pi / mid), type * sin(Pi / mid));
for (int R = mid << 1, j = 0; j < limit; j += R) {
CP w(1, 0);
for (int k = 0; k < mid; k++, w = w * Wn) {
CP x = A[j + k], y = w * A[j + mid + k];
A[j + k] = x + y;
A[j + mid + k] = x - y;
}
}
}
}

void init(int n) {
limit = 1;
while (limit <= n)
limit <<= 1;
for (int i = 1; i < limit; i++)
r[i] = r[i >> 1] >> 1 | ((i & 1) ? limit >> 1 : 0);
}

int MTT(int *a, int *b, int n, int m, int *res, int MOD) {
//	init(n + m);
for (int i = 0; i < n; i++) {
P1[i] = {a[i] / BASE, a[i] % BASE};
P2[i] = {a[i] / BASE, -a[i] % BASE};
}
for (int i = n; i < limit; i++)
P1[i] = P2[i] = {0, 0};
for (int i = 0; i < m; i++)
Q[i] = {b[i] / BASE, b[i] % BASE};
for (int i = m; i < limit; i++)
Q[i] = {0, 0};
FFT(P1, 1);
FFT(P2, 1);
FFT(Q, 1);
for (int i = 0; i < limit; i++) {
Q[i].x /= limit, Q[i].y /= limit;
P1[i] = P1[i] * Q[i], P2[i] = P2[i] * Q[i];
}

FFT(P1, -1);
FFT(P2, -1);
for (int i = 0; i < n + m - 1; i++) {

ll a1b1, a1b2, a2b1, a2b2;

a1b1 = (ll)floor((P1[i].x + P2[i].x) / 2 + 0.5) % MOD;

a1b2 = (ll)floor((P1[i].y + P2[i].y) / 2 + 0.5) % MOD;

a2b1 = (ll)floor((P1[i].y - P2[i].y) / 2 + 0.5) % MOD;

a2b2 = (ll)floor((P2[i].x - P1[i].x) / 2 + 0.5) % MOD;

res[i] = ((a1b1 * BASE + (a1b2 + a2b1)) * BASE + a2b2) % MOD;

res[i] = (res[i] + MOD) % MOD;
}
return n + m - 1;
}

int B[MAXN * 8], tot;

struct Poly {
int *a, len;
void init0(int _len) {
len = _len;
a = B + tot;
for (int i = 0; i < len; i++)
a[i] = 0;
tot += len;
}
void init(int _len, int *src) {
len = _len;
a = B + tot;
for (int i = 0; i < len; i++)
a[i] = src[i];
tot += len;
}
void mul(const Poly &rhs) {
len = MTT(a, rhs.a, len, rhs.len, a, MOD);
}
};

const int LEN = 131072+2;

signed main() {
freopen("dichromatic.in","r",stdin);
freopen("dichromatic.out","w",stdout);
int n, hh;
cin>>n>>hh;
static int F[MAXN], G[MAXN], H[MAXN],tmp[MAXN];
static int ans[MAXN];
init(LEN+LEN);
F[0]=1;
F[1]=1;
for (int t = 1; t <= hh ; t++) {
for (int i = 0; i <= LEN; i++) {
H[i] = (F[i] + G[i]) % MOD;
ans[i] = (ans[i] + H[i]) % MOD;
}
MTT(H, H, LEN, LEN, tmp, MOD);
rep(i,1,LEN) G[i]=tmp[i-1];
MTT(G, G, LEN, LEN, tmp, MOD);
rep(i,1,LEN) F[i]=tmp[i-1];
F[0]=G[0]=0;
}
for (int i = 0; i <= LEN; i++) {
H[i] = (F[i] + G[i]) % MOD;
ans[i] = (ans[i] + H[i]) % MOD;
}
for(int i=1;i<=n;i++){
int xx;
cin>>xx;
cout<<ans[xx]<<" ";
}
}

posted @ 2021-08-23 19:28  GhostCai  阅读(73)  评论(0编辑  收藏  举报