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【CodeForces438E】The Child and Binary Tree

【题目链接】

【思路要点】

  • 按照题解的做法，我们可以通过多项式开根、多项式求逆来求解本题。（但是我不会，也不想学）
  • 我们来考虑一种比较容易的做法，首先：$$dp_s=\sum_{c_i+j+k=s}dp_j*dp_k$$
  • 也就是说，我们只有知道了\(s\)较小的\(dp\)值，才能计算出\(s\)较大的\(dp\)值。
  • 考虑分治，对于区间\([L,R]\)，令\(Mid=\lfloor\frac{L+R}{2}\rfloor\)，我们先计算\(s\in[L,Mid]\)的\(dp\)值，然后考虑\(s\in[L,Mid]\)的\(dp\)值对\(s\in[Mid+1,R]\)的\(dp\)值的影响，再计算\(s\in[Mid+1,R]\)的\(dp\)值。
  • 在考虑\(s\in[L,Mid]\)的\(dp\)值对\(s\in[Mid+1,R]\)的\(dp\)值的影响时，我们需要计算出\(delta_s=\sum_{c_i+j+k=s}dp_j*dp_k(Max\{j,k\}≥L)\)，再将\(delta_s\)加至\(dp_s\)中。
  • 分两种情况，首先，如果\(R≥2*L\)，稍加分析，我们发现，这实际上代表当前处理的节点是分治树上最靠左的节点，换而言之，\(L=0\)，所以，直接NTT卷积三个多项式就可以了。
  • 否则，我们发现，\(Min\{j,k\}<L\)始终成立，因此，我们取\(s\in[L,Mid]\)的\(dp\)值和\(s\in[0,R-L]\)的\(dp\)值卷积，将结果每位乘上2，再和\(c_i\)卷积即可。
  • 时间复杂度\(O(MLog^2M)\)。
  • UPD：现在我学会了多项式开根、多项式求逆，我们来看看怎么解这个题。
  • 令多项式\(F_i=f_i\)，\(C_i=[i\ exist\ in\ c]\)，则有\(F\equiv F*F*C+1(mod\ x^{m+1})\)。
  • 解这个方程，有\(F\equiv \frac{1\pm\sqrt{1-4C}}{2C}(mod\ x^{m+1})\)。
  • 分式上下同乘\(1\mp\sqrt{1-4C}\)，有\(F\equiv \frac{4C}{2C*(1\pm\sqrt{1-4C})}\equiv \frac{2}{1\pm\sqrt{1-4C}}(mod\ x^{m+1})\)。
  • 注意到\(1-4C\)常数项为1，因此\(1-\sqrt{1-4C}\)常数项为0，因此它不存在逆元。
  • 因此，取\(F\equiv \frac{2}{1+\sqrt{1-4C}}(mod\ x^{m+1})\)。
  • 按该式计算\(F\)即可，时间复杂度降至\(O(MLogM)\)。

【代码】

/*Inverse && Sqrt of Polynomial O(NLogN)*/
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 524288;
const int P = 998244353;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
namespace NTT {
	const int G = 3;
	int power(int x, int y) {
		if (y == 0) return 1;
		int tmp = power(x, y / 2);
		if (y % 2 == 0) return 1ll * tmp * tmp % P;
		else return 1ll * tmp * tmp % P * x % P;
	}
	int N, Log, home[MAXN];
	void NTTinit() {
		for (int i = 0; i < N; i++) {
			int ans = 0, tmp = i;
			for (int j = 1; j <= Log; j++) {
				ans <<= 1;
				ans += tmp & 1;
				tmp >>= 1;
			}
			home[i] = ans;
		}
	}
	void NTT(int *a, int mode) {
		for (int i = 0; i < N; i++)
			if (home[i] < i) swap(a[i], a[home[i]]);
		for (int len = 2; len <= N; len <<= 1) {
			int delta;
			if (mode == 1) delta = power(G, (P - 1) / len);
			else delta = power(G, P - 1 - (P - 1) / len);
			for (int i = 0; i < N; i += len) {
				int now = 1;
				for (int j = i, k = i + len / 2; k < i + len; j++, k++) {
					int tmp = a[j];
					int tnp = 1ll * a[k] * now % P;
					a[j] = (tmp + tnp) % P;
					a[k] = (tmp - tnp + P) % P;
					now = 1ll * now * delta % P;
				}
			}
		}
		if (mode == -1) {
			int inv = power(N, P - 2);
			for (int i = 0; i < N; i++)
				a[i] = 1ll * a[i] * inv % P;
		}
	}
	void times(int *a, int *b, int *c, int limit) {
		N = 1, Log = 0;
		while (N < 2 * limit) {
			N <<= 1;
			Log++;
		}
		for (int i = limit; i < N; i++)
			a[i] = b[i] = 0;
		NTTinit();
		NTT(a, 1);
		NTT(b, 1);
		for (int i = 0; i < N; i++)
			c[i] = 1ll * a[i] * b[i] % P;
		NTT(c, -1);
	}
	void timesabb(int *a, int *b, int *c, int limit) {
		N = 1, Log = 0;
		while (N < 2 * limit) {
			N <<= 1;
			Log++;
		}
		for (int i = limit; i < N; i++)
			a[i] = 0;
		for (int i = limit / 2; i <= N; i++)
			b[i] = 0;
		NTTinit();
		NTT(a, 1);
		NTT(b, 1);
		for (int i = 0; i < N; i++)
			c[i] = 1ll * a[i] * b[i] % P * b[i] % P;
		NTT(c, -1);
	}
	void inverse(int *a, int *b, int limit) {
		for (int i = 0; i < 2 * limit; i++) {
			if (i >= limit) a[i] = 0;
			b[i] = 0;
		}
		b[0] = power(a[0], P - 2);
		for (int now = 1; now < limit; now <<= 1) {
			static int c[MAXN], d[MAXN];
			for (int i = 0; i < now * 2; i++)
				c[i] = a[i], d[i] = b[i];
			timesabb(c, d, d, now * 2);
			for (int i = 0; i < now * 2; i++)
				b[i] = (2ll * b[i] - d[i] + P) % P;
		}
	}
	void getsqrt(int *a, int *b, int limit, int residue) {
		for (int i = 0; i < 2 * limit; i++) {
			if (i >= limit) a[i] = 0;
			b[i] = 0;
		}
		b[0] = residue; int inv = power(2, P - 2);
		for (int now = 1; now < limit; now <<= 1) {
			static int c[MAXN], d[MAXN];
			for (int i = 0; i < now * 2; i++)
				c[i] = a[i];
			inverse(b, d, now * 2);
			times(c, d, d, now * 2);
			for (int i = 0; i < now * 2; i++)
				b[i] = 1ll * (b[i] + d[i]) * inv % P;
		}
	}
}
int n, m;
int f[MAXN], c[MAXN], g[MAXN];
int main() {
	read(n), read(m);
	for (int i = 1; i <= n; i++) {
		int x; read(x);
		c[x] = P - 4;
	}
	c[0] = 1;
	NTT::getsqrt(c, g, m + 1, 1);
	g[0] = (g[0] + 1) % P;
	NTT::inverse(g, f, m + 1);
	for (int i = 1; i <= m; i++)
		writeln(f[i] * 2 % P);
	return 0;
}
/*DivideAndConquer + NTT O(NLog^2N)*/
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 262144;
const int P = 998244353;
const int G = 3;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
namespace NTT {
	long long power(int x, int y) {
		if (y == 0) return 1;
		long long tmp = power(x, y / 2);
		if (y % 2 == 0) return tmp * tmp % P;
		else return tmp * tmp % P * x % P;
	}
	int N, Log, home[MAXN];
	void NTTinit() {
		for (int i = 0; i < N; i++) {
			int ans = 0, tmp = i;
			for (int j = 1; j <= Log; j++) {
				ans <<= 1;
				ans += tmp & 1;
				tmp >>= 1;
			}
			home[i] = ans;
		}
	}
	void NTT(long long *a, int mode) {
		for (int i = 0; i < N; i++)
			if (home[i] < i) swap(a[i], a[home[i]]);
		for (int len = 2; len <= N; len <<= 1) {
			long long delta;
			if (mode == 1) delta = power(G, (P - 1) / len);
			else delta = power(G, P - 1 - (P - 1) / len);
			for (int i = 0; i < N; i += len) {
				long long now = 1;
				for (int j = i, k = i + len / 2; k < i + len; j++, k++) {
					long long tmp = a[j];
					long long tnp = a[k] * now % P;
					a[j] = (tmp + tnp) % P;
					a[k] = (tmp - tnp + P) % P;
					now = now * delta % P;
				}
			}
		}
		if (mode == -1) {
			long long inv = power(N, P - 2);
			for (int i = 0; i < N; i++)
				a[i] = a[i] * inv % P;
		}
	}
	void times(long long *a, long long *b, long long *c, int limit) {
		N = 1, Log = 0;
		while (N <= 2 * limit) {
			N <<= 1;
			Log++;
		}
		for (int i = limit; i < N; i++)
			a[i] = b[i] = 0;
		NTTinit();
		NTT(a, 1);
		NTT(b, 1);
		for (int i = 0; i < N; i++)
			c[i] = a[i] * b[i] % P;
		NTT(c, -1);
		for (int i = limit; i < N; i++)
			c[i] = 0;
	}
};
bool exist[MAXN];
long long dp[MAXN];
void solve(int l, int r) {
	if (l == r) return;
	int mid = (l + r) / 2;
	solve(l, mid);
	int len = r - l;
	static long long a[MAXN], b[MAXN], c[MAXN];
	if (len >= l) {
		for (int i = 0; i <= r; i++)
			a[i] = b[i] = dp[i];
		NTT::times(a, b, c, r + 1);
		for (int i = 0; i <= r; i++)
			b[i] = exist[i];
		NTT::times(c, b, a, r + 1);
		for (int i = mid + 1; i <= r; i++)
			dp[i] = (dp[i] + a[i]) % P;
	} else {
		for (int i = l; i <= mid; i++)
			a[i - l] = dp[i];
		for (int i = mid + 1; i <= r; i++)
			a[i - l] = 0;
		for (int i = 0; i <= len; i++)
			b[i] = dp[i];
		NTT::times(a, b, c, len + 1);
		for (int i = 0; i <= len; i++) {
			c[i] = c[i] * 2 % P;
			b[i] = exist[i];
		}
		NTT::times(c, b, a, len + 1);
		for (int i = mid + 1; i <= r; i++)
			dp[i] = (dp[i] + a[i - l]) % P;
	}
	solve(mid + 1, r);
}
int main() {
	int n, m;
	read(n), read(m);
	for (int i = 1; i <= n; i++) {
		int x; read(x);
		exist[x] = true;
	}
	dp[0] = 1;
	solve(0, m);
	for (int i = 1; i <= m; i++)
		writeln(dp[i]);
	return 0;
}