朱老师的难题

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朱老师的难题

题目大意

算法

设多项式$G(x)$，$x^n$的系数表示一组选$n$个的乘积之和

那么$G(x)=1-\prod_{i=1}^n (1-A_ix)$

设多项式$F(x)$，$x^n$的系数表示一套选$n$个的答案

那么

要把$[x^n]F(x)$写成$\sum_{i=1}^n B_iC_i^n$的形式

构造：$F(x)=\sum_{i=1}^n \frac {a_i} {1-A_ix}$，这样$[x^n]F(x) = \sum_{i=1}^n a_i A_i^n$

考虑如何求出$a_i$

设多项式$H(x) = \sum_{i=1}^n\prod_{j!=i}(1-A_jx)$

那么$\frac 1 {a_i}=H(\frac 1 {A_i})$

$H(x)$可以分治求得，再上一个多项式多点求值就好了

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#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
const int N = 1 << 18, mod = 1811939329;

char buf[1 << 20], *p1, *p2;
#define GC (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2) ? 0 : *p1 ++)
inline int _R() {
	int d = 0; bool ty = 1; char t;
	while (t = GC, (t < '0' || t > '9') && t != '-');
	t == '-' ? (ty = 0) : (d = t - '0');
	while (t = GC, t >= '0' && t <= '9') d = (d << 3) + (d << 1) + t - '0';
	return ty ? d : -d;
}
inline int add(int a, int b) { long long c = (long long) a + b; return c >= mod ? c - mod : c; }
inline int sub(int a, int b) { a -= b; return a < 0 ? a + mod : a; }
inline void inc(int &a, int b) { a = add(a, b); }
inline int mul(int a, int b) { return 1LL * a * b % mod; }
inline int ksm(int a, int b) { int r; for (r = 1; b; b >>= 1, a = mul(a, a)) if (b & 1) r = mul(r, a); return r; }

int ntt_wi[2][N];
void pre_ntt() {
	int f = ksm(13, (mod - 1) / N), g = ksm(f, mod - 2);
	ntt_wi[0][0] = ntt_wi[1][0] = 1;
	for (int i = 1; i < N; i ++) {
		ntt_wi[1][i] = mul(ntt_wi[1][i - 1], f);
		ntt_wi[0][i] = mul(ntt_wi[0][i - 1], g);
	}
}

void ntt(int A[], int n, int ty) {
	int i, j, k, t, w; int f, g;
	for (i = j = 0; i < n; i ++) {
		if (i < j) swap(A[i], A[j]);
		for (k = n >> 1; (j ^= k) < k; k >>= 1);
	}
	for (w = 1; t = N / (w << 1), w < n; w <<= 1)
		for (k = 0; k < n; k += w << 1)
			for (i = k, j = 0; i < k + w; i ++, j += t) {
				f = A[i], g = mul(A[i + w], ntt_wi[ty][j]);
				A[i] = add(f, g), A[i + w] = sub(f, g);
			}
	if (ty == 1) return;
	f = ksm(n, mod - 2);
	for (i = 0; i < n; i ++) A[i] = mul(A[i], f);
}

void poly_inverse(int n, int A[], int B[]) {
	static int AA[N];
	if (n == 1) {
		B[0] = ksm(A[0], mod - 2);
		return;
	}
	int i, half = n >> 1, p = n << 1;
	poly_inverse(half, A, B);
	
	fill(B + half, B + p, 0);
	copy(A, A + n, AA);
	fill(AA + n, AA + p, 0);
	
	ntt(B, p, 1), ntt(AA, p, 1);
	for (i = 0; i < p; i ++) B[i] = mul(B[i], sub(2, mul(B[i], AA[i])));
	ntt(B, p, 0);
}

struct poly : vector <int> {
	poly() {}
	poly(const vector <int>& cp) : vector <int> (cp) {}
	int deg() const { return size() - 1; }
	poly& operator %= (const poly& o);
} ;

poly operator * (const poly& lhs, const poly& rhs) {
	static int A[N], B[N];
	int i, p, n = lhs.deg() + rhs.deg();
	for (p = 1; p <= n; p <<= 1);
	copy(lhs.begin(), lhs.end(), A);
	copy(rhs.begin(), rhs.end(), B);
	fill(A + lhs.deg() + 1, A + p, 0);
	fill(B + rhs.deg() + 1, B + p, 0);
	ntt(A, p, 1), ntt(B, p, 1);
	for (i = 0; i < p; i ++) A[i] = mul(A[i], B[i]);
	ntt(A, p, 0);
	return vector <int> (A, A + n + 1);
}

poly& poly :: operator %= (const poly& A) {
	static int tmp[N], A0[N];
	if (this -> deg() < A.deg()) return *this;

	int i, n, p, m = deg() - A.deg() + 1;
	for (p = 1; p < m << 1; p <<= 1);
	reverse_copy(A.begin(), A.end(), tmp);
	fill(tmp + m, tmp + p, 0);
	fill(A0, A0 + p, 0);
	poly_inverse(p >> 1, tmp, A0);
	fill(A0 + m, A0 + p, 0);

	reverse_copy(begin(), end(), tmp);
	fill(tmp + min(m, deg() + 1), tmp + p, 0);
	ntt(tmp, p, 1), ntt(A0, p, 1);
	for (i = 0; i < p; i ++) tmp[i] = mul(tmp[i], A0[i]);
	ntt(tmp, p, 0);

	poly mult = vector <int> (tmp, tmp + m);
	reverse(mult.begin(), mult.end());
	mult = mult * A, n = deg(), m = 0;
	for (i = 0; i < n; i ++) {
		(*this) [i] = sub((*this) [i], mult[i]);
		if ((*this) [i]) m = i;
	}
	erase(begin() + m + 1, end());
	return *this;
}


int n, A[N], H[N];
int totn, ls[N], rs[N], B[N], C[N];
poly f, P[N];

void dc(int d, int l, int r) {
	static int AAA[20][N], HH[20][N], AA[N];
	if (l == r) { AA[0] = 1, AA[1] = mod - A[l], H[0] = 1; return; }
	int i, mid = l + r >> 1, len = r - l + 1, llen = mid - l + 1, p;

	dc(d + 1, l, mid);
	for (p = 1; p <= (len << 1); p <<= 1);
	copy(H, H + llen, HH[d]);
	fill(HH[d] + llen, HH[d] + p, 0);
	fill(H, H + p, 0);
	copy(AA, AA + llen + 1, AAA[d]);
	fill(AAA[d] + llen + 1, AAA[d] + p, 0);
	fill(AA + 0, AA + p, 0);
	dc(d + 1, mid + 1, r);

	ntt(HH[d], p, 1), ntt(H, p, 1), ntt(AAA[d], p, 1), ntt(AA, p, 1);
	for (i = 0; i < p; i ++) H[i] = add(mul(H[i], AAA[d][i]), mul(HH[d][i], AA[i])), AA[i] = mul(AAA[d][i], AA[i]);
	ntt(H, p, 0), ntt(AA, p, 0);
}

void getp(int &p, int l, int r) {
	p = ++ totn;
	if (l < r) {
		int mid = l + r >> 1;
		getp(ls[p], l, mid);
		getp(rs[p], mid + 1, r);
		P[p] = P[ls[p]] * P[rs[p]];
	} else P[p] = vector <int> {mod - B[l], 1};
}
void getans(int p, int l, int r, poly x) {
	x %= P[p];
	if (l == r) { C[l] = x[0]; return; }
	int mid = l + r >> 1;
	getans(ls[p], l, mid, x);
	getans(rs[p], mid + 1, r, x);
}

int main() {
	int i;
	pre_ntt();
	n = _R();
	for (i = 1; i <= n; i ++) A[i] = _R();

	dc(0, 1, n);
	for (i = 0; i < n; i ++) f.push_back(H[i]);

	for (i = 1; i <= n; i ++) B[i] = ksm(A[i], mod - 2);
	getp(i, 1, n);
	getans(1, 1, n, f);

	for (i = 1; i <= n; i ++) printf("%d %d\n", ksm(C[i], mod - 2), A[i]);
}