RXD and functions HDU-6061 NTT

RXD and functions

solution

gm=f(x−∑i=1mai),令A=∑i=1mai，则有，g_m=f(x-\sum\limits_{i=1}^{m}a_i),令A=\sum\limits_{i=1}^{m}a_i，则有，gm=f(x−i=1∑mai),令A=i=1∑mai，则有，

gm=f(x−A)g_m=f(x-A)gm=f(x−A)

=∑i=0nci(x−a)i=\sum\limits_{i=0}^{n}c_i(x-a)^i=i=0∑nci(x−a)i

=∑i=0nci∑j=0iCij(−a)i−jxj=\sum\limits_{i=0}^{n}c_i\sum\limits_{j=0}^{i}C_i^j(-a)^{i-j}x^j=i=0∑ncij=0∑iCij(−a)i−jxj

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−-----------------------------−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
c0[C00(−a)0x0]c_0[C_0^0(-a)^0x^0]c0[C00(−a)0x0]

c1[C10(−a)1x0+C11(−a)0x1]c_1[C_1^0(-a)^1x^0+C_1^1(-a)^0x^1]c1[C10(−a)1x0+C11(−a)0x1]

c2[C20(−a)2x0+C21(−a)1x1+C22(−a)0x2]c_2[C_2^0(-a)^2x^0+C_2^1(-a)^1x^1+C_2^2(-a)^0x^2]c2[C20(−a)2x0+C21(−a)1x1+C22(−a)0x2]
…
cn[Cn0(−a)nx0+Cn1(−a)n−1x1+Cn0(−a)n−2x2+...+Cnn(−a)0xn]c_n[C_n^0(-a)^nx^0+C_n^1(-a)^{n-1}x^1+C_n^0(-a)^{n-2}x^2+...+C_n^n(-a)^{0}x^n]cn[Cn0(−a)nx0+Cn1(−a)n−1x1+Cn0(−a)n−2x2+...+Cnn(−a)0xn]

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−-----------------------------−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

=∑j=0n∑i=jnci+j(−a)i−jCijxj=\sum\limits_{j=0}^{n}\sum\limits_{i=j}^{n}c_{i+j}(-a)^{i-j}C_i^jx^j=j=0∑ni=j∑nci+j(−a)i−jCijxj

=∑j=0n∑i=0n−jci+j(−a)iCi+jjxj=\sum\limits_{j=0}^{n}\sum\limits_{i=0}^{n-j}c_{i+j}(-a)^{i}C_{i+j}^jx^j=j=0∑ni=0∑n−jci+j(−a)iCi+jjxj

=∑j=0n∑i=0n−jci+j(−a)i(i+j)!i!j!xj=\sum\limits_{j=0}^{n}\sum\limits_{i=0}^{n-j}c_{i+j}(-a)^{i}\frac{(i+j)!}{i!j!}x^j=j=0∑ni=0∑n−jci+j(−a)ii!j!(i+j)!xj

=∑j=0nbjxj=\sum\limits_{j=0}^{n}b_jx^j=j=0∑nbjxj

=>=>=>

∑j=0nj!bjxj=∑j=0n∑i=0n−j(i+j)!ci+j(−a)ii!xj\sum\limits_{j=0}^{n}j!b_jx^j=\sum\limits_{j=0}^{n}\sum\limits_{i=0}^{n-j}(i+j)!c_{i+j}\frac{(-a)^{i}}{i!}x^jj=0∑nj!bjxj=j=0∑ni=0∑n−j(i+j)!ci+ji!(−a)ixj

=>=>=>

j!bj=∑i=0n−j(i+j)!ci+j∗(−a)ii!j!b_j=\sum\limits_{i=0}^{n-j}(i+j)!c_{i+j}*\frac{(-a)^{i}}{i!}j!bj=i=0∑n−j(i+j)!ci+j∗i!(−a)i

令F[j]=(n−j)!cn−j，G[j]=(−a)jj!令F[j]={(n-j)}!c_{n-j}，G[j]=\frac{(-a)^{j}}{j!}令F[j]=(n−j)!cn−j，G[j]=j!(−a)j

=>=>=>

∑j=0nj!bj=∑j=0n∑i=0jF[j−i]G[i]\sum\limits_{j=0}^{n}j!b_j=\sum\limits_{j=0}^{n}\sum\limits_{i=0}^{j}F[j-i]G[i]j=0∑nj!bj=j=0∑ni=0∑jF[j−i]G[i]

通过F∗G得到Gm.通过F*G得到G_m.通过F∗G得到Gm.

code

/*SiberianSquirrel*//*CuteKiloFish*/
#include <bits/stdc++.h>
using namespace std;
#define gcd(a,b) __gcd(a,b)
#define Polynomial vector<int>
#define Inv(x) quick_pow(x, mod - 2)
using ll = long long;
using ull = unsigned long long;
const ll mod = 998244353, mod_g = 3, img = 86583718;//NTT, 多项式三角形
const int N = int(6e5 + 10);inline int add(int a, int b) {return a + b < mod? a + b: a + b - mod;
}
inline int sub(int a, int b) {return a - b < 0? a - b + mod: a - b;
}
inline int mul(int a, int b) {return 1ll * a * b % mod;
}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(register int i = 0; i < n - 1; ++ i) Cinv[i] = Cinv[i + 1]; Cinv[n - 1] = 0;Cinv = Inverse(Cinv);for(register 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;
}int fac[int(1e5 + 10)] = {1}, ifac[int(1e5 + 10)] = {1};inline void solve() {for(int i = 1; i <= int(1e5); ++ i) fac[i] = 1ll * fac[i - 1] * i % mod, ifac[i] = Inv(fac[i]);int n, m;while(cin >> n) {vector<int> c(n + 1);for(int i = 0; i <= n; ++ i) cin >> c[i];cin >> m;int A = 0;for(int i = 1; i <= m; ++ i) {int x; cin >> x;A = add(A, x);}Polynomial F(n + 1), G(n + 1);int temp = 1;for(int i = 0; i <= n; ++ i) {F[i] = 1ll * fac[n - i] * c[n - i] % mod;G[i] = 1ll * temp * ifac[i] % mod;temp = mod - 1ll * temp * A % mod;}Polynomial B = F * G;for(int i = 0; i <= n; ++ i) {cout << 1ll * B[n - i] * ifac[i] % mod << ' ';}cout << '\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;
}