无限手套 分治 + NTT + 生成函数

无限手套

solution

$$\prod _{k=0}^{\infty }{a}_i{x}_i^2+b_i{x}_i+1$$
$考虑生成函数：$
$$=>\sum _{k=0}^{\infty }{a}_i{k}^2{x}^k+b_i{x}_i{x}^k+x^k$$$$=\sum _{k=0}^{\infty }{a}_i{k}^2{x}^k+\sum _{k=0}^{\infty }{b}_i{x}_i{x}^k+\sum _{k=0}^{\infty }{x}^k,(三项分别当作A,B,C)$$
$$C=\sum _{k=0}^{\infty }{x}^k=\frac{1}{1-x}$$

$B=\sum _{k=0}^{\infty }{b}_i{x}_i{x}^k$
$xB=\sum _{k=1}^{\infty }{b}_i{x}_i{x}^{k+1}，两式联立解得：$
$$B=\frac{b_{i}x}{(1-x{)}^2}$$

$A=\sum _{k=0}^{\infty }{a}_i{k}^2{x}^k$
$C=\sum _{k=0}^{\infty }{x}^k=\frac{1}{1-x}$
$C^{\prime }=\sum _{k=0}^{\infty }k{x}^{k-1}=\frac{1}{(1-x{)}^2}$
$x{C}^{\prime }=\sum _{k=0}^{\infty }k{x}^k=\frac{x}{(1-x{)}^2}$
$(x{C}^{\prime }{)}^{\prime }=\sum _{k=0}^{\infty }{k}^2{x}^{k-1}=\frac{(x+1)}{(1-x{)}^3}$
$x(x{C}^{\prime }{)}^{\prime }=\sum _{k=0}^{\infty }{k}^2{x}^k=\frac{x(x+1)}{(1-x{)}^3}$
$$A=a_i\ast x(x{C}^{\prime }{)}^{\prime }=\frac{a_{i}x(x+1)}{(1-x{)}^3}$$

$因此，$

$$\sum _{k=0}^{\infty }{a}_i{k}^2{x}^k+b_i{x}_i{x}^k+x^k$$$$=\frac{a_{i}x(x+1)}{(1-x{)}^3}+\frac{b_{i}x}{(1-x{)}^2}+\frac{1}{1-x}$$$$=\frac{a_{i}x(x+1)+(1-x)b_{i}x+(1-x{)}^2}{(1-x{)}^3}$$$$=\frac{1+(a_i+b_i-2)x+(a_i-b_i+1)x^2}{(1-x{)}^3}$$

$$Answer=\prod _{i=1}^m\frac{1+(a_i+b_i-2)x+(a_i-b_i+1)x^2}{(1-x{)}^3}$$$$=\frac{\prod _{i=1}^{m}1+(a_i+b_i-2)x+(a_i-b_i+1)x^2}{(1-x{)}^{3m}}$$

$$\frac{1}{(1-x{)}^n}=\sum _{i=0}^{\infty }{C}_{n+i-1}^i{x}^i$$

$$Answer=\prod _{i=1}^m(1+(a_i+b_i-2)x+(a_i-b_i+1)x^2)\ast \sum _{i=0}^{\infty }{C}_{3m+i-1}^i{x}^i$$

由于这个式子的复杂度会达到$O(mnlogn<2e8),m<1e3,n<1e4,logn<20$,因此需要运用分治。

code

/*SiberianSquirrel*//*CuteKiloFish*/
#include <bits/stdc++.h>
using namespace std;
#define gcd(a,b) __gcd(a,b)
#define Polynomial vector<ll>
#define Inv(x) quick_pow(x, mod - 2)
#define DEBUG(x, y) cout << x << ": " << y << '\n';
using ld = long double;
using ll = long long;
using ull = unsigned long long;
const ll mod = 998244353, mod_g = 3, img = 86583718;
//const ll mod = 1004535809, mod_g = 3;
const int N = int(1e5 + 10);
template<typename T> inline T read() {T x = 0, f = 1;char ch = getchar();while(ch < '0' || ch > '9') { if(ch == '-') f = -1; ch = getchar(); }while(ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }return x * f;
}
template<typename T> inline T print(T x) {if(x < 0) { putchar('-'); x =- x; }if(x > 9) print(x / 10);putchar(x % 10 + '0');
}void print(Polynomial &a, int len){for(int i = 0; i < len; ++ i)cout << a[i] << ' ';cout << '\n';
}
void exgcd(int a,int b,int &x,int &y){if(!b) {x = 1; y = 0;return;}exgcd(b, a % b, y, x);y -= x * (a / b);
}
int quick_pow(int ans, int p, int res = 1) {for(; p; p >>= 1, ans = 1ll * ans * ans % mod)if(p & 1) res = 1ll * res * ans % mod;return res % mod;
}Polynomial R;
//二进制向上取整，为方便NTT变换准备。
inline int Binary_Rounding(const int &n) {int len = 1; while(len < n) len <<= 1;return len;
}
//预处理R数组，准备变换，在每次NTT之前理论都要调用此函数。
inline int Prepare_Transformation(int n){int L = 0, len;for(len = 1; len < n; len <<= 1) L++;R.clear(); R.resize(len);for(int i = 0; i < len; ++ i)R[i] = (R[i>>1]>>1)|((i&1)<<(L-1));return len;
}
void NTT(Polynomial &a, int f){int n = a.size();for(int i = 0; i < n; ++ i)if(i < R[i])swap(a[i], a[R[i]]);for(int i = 1; i < n; i <<= 1)for(int j = 0, gn = quick_pow(mod_g,(mod - 1) / (i<<1)); j < n; j += (i<<1))for(int k = 0, g = 1, x, y; k < i; ++ k, g = 1ll * g * gn % mod)x = a[j + k], y = 1ll * g * a[i + j + k] % mod,a[j + k] = (x + y) % mod, a[i + j + k] = (x - y + mod) % mod;if(f == -1){reverse(a.begin() + 1, a.end());int inv = Inv(n);for(int i = 0; i < n; ++ i) a[i] = 1ll * a[i] * inv % mod;}
}inline Polynomial operator +(const Polynomial &a, const int &b){int sizea = a.size(); Polynomial ret = a; ret.resize(sizea);for(int i = 0; i < sizea; ++ i)ret[i] = (1ll * a[i] + b + mod) % mod;return ret;
}
inline Polynomial operator -(const Polynomial &a, const int &b){int sizea = a.size(); Polynomial ret = a; ret.resize(sizea);for(int i = 0; i < sizea; ++ i)ret[i] = (1ll * a[i] - b + mod) % mod;return ret;
}
inline Polynomial operator *(const Polynomial &a, const int &b){int sizea = a.size(); Polynomial ret = a; ret.resize(sizea);for(int i = 0; i < sizea; ++ i) ret[i] = (1ll * a[i] * b % mod + mod) % mod;return ret;
}
inline Polynomial operator +(const Polynomial &a, const Polynomial &b){int sizea = a.size(), sizeb = b.size(), size = max(sizea, sizeb);Polynomial ret = a; ret.resize(size);for(int i = 0; i < sizeb; ++ i) ret[i] = (1ll * ret[i] + b[i]) % mod;return ret;
}
inline Polynomial operator -(const Polynomial &a, const Polynomial &b){int sizea = a.size(), sizeb = b.size(), size = max(sizea, sizeb);Polynomial ret = a; ret.resize(size);for(int i = 0; i < sizeb; ++ i) ret[i] = (1ll * ret[i] - b[i] + mod) % mod;return ret;
}
inline Polynomial Inverse(const Polynomial &a){Polynomial ret, inv_a;ret.resize(1);ret[0] = Inv(a[0]); int ed = a.size();for(int len = 2; len <= ed; len <<= 1){int n = Prepare_Transformation(len << 1);inv_a = a; inv_a.resize(n); ret.resize(n);for(int i = len; i < n; ++ i) inv_a[i] = 0;NTT(inv_a, 1); NTT(ret, 1);for(int i = 0; i < n; ++ i)ret[i] = 1ll * (2ll - 1ll * inv_a[i] * ret[i] % mod + mod) % mod * ret[i] % mod;NTT(ret, -1);for(int i = len; i < n; ++ i) ret[i] = 0;}ret.resize(ed);return ret;
}
inline Polynomial operator *(const Polynomial &a, const Polynomial &b){Polynomial lsa = a, lsb = b, ret;int n = lsa.size(), m = lsb.size();n = Prepare_Transformation(n + m);lsa.resize(n); lsb.resize(n); ret.resize(n);NTT(lsa,1); NTT(lsb,1);for(int i = 0; i < n; ++ i) ret[i] = 1ll * lsa[i] * lsb[i] % mod;NTT(ret,-1);return ret;
}
inline Polynomial operator /(const Polynomial &a, const Polynomial &b){Polynomial ret = a, ls = b;reverse(ret.begin(), ret.end());reverse(ls.begin(), ls.end());ls.resize(Binary_Rounding(a.size() + b.size()));ls = Inverse(ls);ls.resize(a.size() + b.size());ret = ret * ls; ret.resize(a.size() - b.size() + 1);reverse(ret.begin(), ret.end());return ret;
}
inline Polynomial operator %(const Polynomial &a, const Polynomial &b){Polynomial ret = a / b;ret = ret * b; ret.resize(a.size() + b.size());ret = a - ret; ret.resize(a.size() + b.size());return ret;
}
inline Polynomial Derivation(const Polynomial &a){int size = a.size(); Polynomial ret; ret.resize(size);for(int i = 1; i < size; ++ i) ret[i - 1] = 1ll * i * a[i] % mod;ret[size - 1] = 0;return ret;
}
inline Polynomial Integral(const Polynomial &a){int size = a.size(); Polynomial ret; ret.resize(size);for(int i = 1; i < size; ++ i) ret[i] = 1ll * Inv(i) * a[i - 1] % mod;ret[0] = 0;return ret;
}
inline Polynomial Composition_Inverse(const Polynomial &a){int n = a.size();Polynomial ret, Cinv = a, Pow;Cinv.resize(n); ret.resize(n); Pow.resize(n); Pow[0] = 1;for(int i = 0; i < n - 1; ++ i) Cinv[i] = Cinv[i + 1]; Cinv[n - 1] = 0;Cinv = Inverse(Cinv);for(int i = 1; i < n; ++ i){Pow = Pow * Cinv; Pow.resize(n);ret[i] = 1ll * Pow[i - 1] * Inv(i) % mod;}return ret;
}
inline Polynomial Logarithmic(const Polynomial &a){Polynomial ln_a = Derivation(a) * Inverse(a);ln_a.resize(a.size());return Integral(ln_a);
}
inline Polynomial Exponential(const Polynomial &a, int Constant = 1){Polynomial ret, D; int ed = a.size();ret.resize(1); ret[0] = Constant;for(int len = 2; len <= ed; len <<= 1){D = Logarithmic(ret); D.resize(len);D[0] = (1ll * a[0] + 1ll - D[0] + mod) % mod;for(int i = 1; i < len; ++i) D[i] = (1ll * a[i] - D[i] + mod) % mod;int n = Prepare_Transformation(len<<1);ret.resize(n); D.resize(n);NTT(ret, 1); NTT(D,1);for(int i = 0; i < n; ++ i) ret[i] = 1ll * ret[i] * D[i] % mod;NTT(ret, -1);for(int i = len; i < (len<<1); ++ i) ret[i] = D[i] = 0;}ret.resize(ed);return ret;
}ll fac[int(6e5 + 10)] = {1}, ifac[int(6e5 + 10)] = {1};
ll C(int n, int m) {if(m == 0 || n == m) return 1;if(n < m) return 0;return 1ll * fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}//Polynomial ans(3);
Polynomial a(1010), b(1010);Polynomial DAC_NTT(int l, int r) {if (l == r) {Polynomial res;ll A = 1;ll B = (a[l] + b[l] - 2) % mod; while(B < 0) B += mod;ll C = (a[l] - b[l] + 1) % mod; while(C < 0) C += mod;res.push_back(A); res.push_back(B); res.push_back(C);return res;}int mid = l + r >> 1;return DAC_NTT(l, mid) * DAC_NTT(mid + 1, r);
}inline void solve() {for(int i = 1; i <= int(6e5); ++ i) fac[i] = 1ll * fac[i - 1] * i % mod, ifac[i] = Inv(fac[i]);int m; cin >> m;Polynomial res(50010);for(int i = 0; i <= 50000; ++ i) {res[i] = C(3 * m + i - 1, i);}
//     Polynomial ans;for(int i = 1; i <= m; ++ i) {cin >> a[i] >> b[i];
//         ll A = 1;
//         ll B = (a[i] + b[i] - 2) % mod; while(B < 0) B += mod;
//         ll C = (a[i] - b[i] + 1) % mod; while(C < 0) C += mod;
//         if(i == 1) {
//             ans.push_back(A), ans.push_back(B), ans.push_back(C);
//         } else {
//             Polynomial temp;
//             temp.push_back(A), temp.push_back(B), temp.push_back(C);
//             ans = ans * temp;
//         }}Polynomial ans = DAC_NTT(1, m);ans = ans * res;int q; cin >> q; while(q --) {int n; cin >> n;cout << ans[n] << '\n';}
}signed main() {ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#ifdef ACM_LOCALfreopen("input", "r", stdin);
//    freopen("output", "w", stdout);signed test_index_for_debug = 1;char acm_local_for_debug = 0;do {if (acm_local_for_debug == '$') exit(0);if (test_index_for_debug > 20)throw runtime_error("Check the stdin!!!");auto start_clock_for_debug = clock();solve();auto end_clock_for_debug = clock();cout << "Test " << test_index_for_debug << " successful" << endl;cerr << "Test " << test_index_for_debug++ << " Run Time: "<< double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;cout << "--------------------------------------------------" << endl;} while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
#elsesolve();
#endifreturn 0;
}