tokitsukaze and Another Protoss and Zerg

考慮生成函數，每一場的生成函數為$\sum _{j=1}^{b[i]}{C}_{b[i]}^j+\sum _{j=1}^{a[i]}{C}_{a[i]}^j{x}^j$，

進一步化簡可得$2^{b[i]}?1+\sum _{j=1}^{a[i]}{C}_{a[i]}^j{x}^j$，

把$n$場的全部乘起來，然后輸出第$[x^i]$項的系數就是答案了。

#include <bits/stdc++.h>using namespace std;const int mod = 998244353, inv2 = mod + 1 >> 1;namespace Quadratic_residue {struct Complex {int r, i;Complex(int _r = 0, int _i = 0) : r(_r), i(_i) {}};int I2;Complex operator * (const Complex &a, Complex &b) {return Complex((1ll * a.r * b.r % mod + 1ll * a.i * b.i % mod * I2 % mod) % mod, (1ll * a.r * b.i % mod + 1ll * a.i * b.r % mod) % mod);}Complex quick_pow(Complex a, int n) {Complex ans = Complex(1, 0);while (n) {if (n & 1) {ans = ans * a;}a = a * a;n >>= 1;}return ans;}int get_residue(int n) {mt19937 e(233);if (n == 0) {return 0;}if(quick_pow(n, (mod - 1) >> 1).r == mod - 1) {return -1;}uniform_int_distribution<int> r(0, mod - 1);int a = r(e);while(quick_pow((1ll * a * a % mod - n + mod) % mod, (mod - 1) >> 1).r == 1) {a = r(e);}I2 = (1ll * a * a % mod - n + mod) % mod;int x = quick_pow(Complex(a, 1), (mod + 1) >> 1).r, y = mod - x;if(x > y) swap(x, y);return x;} }const int N = 1e6 + 10;int r[N], inv[N], b[N], c[N], d[N], e[N], t[N];int quick_pow(int a, int n) {int ans = 1;while (n) {if (n & 1) {ans = 1ll * a * ans % mod;}a = 1ll * a * a % mod;n >>= 1;}return ans; }void get_r(int lim) {for (int i = 0; i < lim; i++) {r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);} }void get_inv(int n) {inv[1] = 1;for (int i = 2; i <= n; i++) {inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;} }void NTT(int *f, int lim, int rev) {for (int i = 0; i < lim; i++) {if (i < r[i]) {swap(f[i], f[r[i]]);}}for (int mid = 1; mid < lim; mid <<= 1) {int wn = quick_pow(3, (mod - 1) / (mid << 1));for (int len = mid << 1, cur = 0; cur < lim; cur += len) {int w = 1;for (int k = 0; k < mid; k++, w = 1ll * w * wn % mod) {int x = f[cur + k], y = 1ll * w * f[cur + mid + k] % mod;f[cur + k] = (x + y) % mod, f[cur + mid + k] = (x - y + mod) % mod;}}}if (rev == -1) {int inv = quick_pow(lim, mod - 2);reverse(f + 1, f + lim);for (int i = 0; i < lim; i++) {f[i] = 1ll * f[i] * inv % mod;}} }void polyinv(int *f, int *g, int n) {if (n == 1) {g[0] = quick_pow(f[0], mod - 2);return ;}polyinv(f, g, n + 1 >> 1);for (int i = 0; i < n; i++) {t[i] = f[i];}int lim = 1;while (lim < 2 * n) {lim <<= 1;}get_r(lim);NTT(t, lim, 1);NTT(g, lim, 1);for (int i = 0; i < lim; i++) {int cur = (2 - 1ll * g[i] * t[i] % mod + mod) % mod;g[i] = 1ll * g[i] * cur % mod;t[i] = 0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;} }void polysqrt(int *f, int *g, int n) {if (n == 1) {g[0] = Quadratic_residue::get_residue(f[0]);return ;}polysqrt(f, g, n + 1 >> 1);polyinv(g, b, n);int lim = 1;while (lim < 2 * n) {lim <<= 1;}get_r(lim);for (int i = 0; i < n; i++) {t[i] = f[i];}NTT(g, lim, 1);NTT(b, lim, 1);NTT(t, lim, 1);for (int i = 0; i < lim; i++) {g[i] = (1ll * inv2 * g[i] % mod + 1ll * inv2 * b[i] % mod * t[i] % mod) % mod;b[i] = t[i] = 0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;} }void derivative(int *a, int *b, int n) {for (int i = 0; i < n; i++) {b[i] = 1ll * a[i + 1] * (i + 1) % mod;} }void integrate(int *a, int n) {for (int i = n - 1; i >= 1; i--) {a[i] = 1ll * a[i - 1] * inv[i] % mod;}a[0] = 0; }void polyln(int *f, int *g, int n) {polyinv(f, b, n);derivative(f, g, n);int lim = 1;while (lim < 2 * n) {lim <<= 1;}get_r(lim);NTT(g, lim, 1);NTT(b, lim, 1);for (int i = 0; i < lim; i++) {g[i] = 1ll * g[i] * b[i] % mod;b[i] = 0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;}integrate(g, n); }void polyexp(int *f, int *g, int n) {if (n == 1) {g[0] = 1;return ;}polyexp(f, g, n + 1 >> 1);int lim = 1;while (lim < 2 * n) {lim <<= 1;}polyln(g, d, n);for (int i = 0; i < n; i++) {t[i] = (f[i] - d[i] + mod) % mod;}t[0] = (t[0] + 1) % mod;get_r(lim);NTT(g, lim, 1);NTT(t, lim, 1);for (int i = 0; i < lim; i++) {g[i] = 1ll * g[i] * t[i] % mod;t[i] = d[i] = 0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;} }/*b存放多項式逆，c存放多項式開根，d存放多項式對數ln，e存放多項式指數exp，t作為中間轉移數組,如果要用到polyinv，得提前調用get_inv(n)先預先得到我們想要得到的逆元范圍。 */int fac[N], p[N];int C(int n, int m) {if(n < 0 || m < 0 || m > n) return 0;if(m == 0 || m == n) return 1;return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod; }void init() {fac[0] = p[0] = 1;for(int i = 1; i < N; i++)fac[i] = 1ll * fac[i - 1] * i % mod, p[i] = 1ll * p[i - 1] * 2 % mod;inv[N - 1] = quick_pow(fac[N - 1], mod - 2);for(int i = N - 2; i >= 0; i--)inv[i] = 1ll * inv[i + 1] * (i + 1) % mod; }int A[N], B[N], x[N], y[N], n;vector<int> f[N];void solve(int rt, int l, int r) {if (l == r) {f[rt].push_back((p[B[l]] - 1 + mod) % mod);for (int i = 1; i <= A[l]; i++) {f[rt].push_back(C(A[l], i));}return ;}int mid = l + r >> 1;solve(rt << 1, l, mid);solve(rt << 1 | 1, mid + 1, r);for (int i = 0; i < f[rt << 1].size(); i++) {x[i] = f[rt << 1][i];}for (int i = 0; i < f[rt << 1 | 1].size(); i++) {y[i] = f[rt << 1 | 1][i];}int lim = 1, len = f[rt << 1].size() + f[rt << 1 | 1].size() - 1;while (lim <= len) {lim <<= 1;}get_r(lim);NTT(x, lim, 1);NTT(y, lim, 1);for (int i = 0; i < lim; i++) {x[i] = 1ll * x[i] * y[i] % mod;}NTT(x, lim, -1);for (int i = 0; i <= len; i++) {f[rt].push_back(x[i]);}for (int i = 0; i < lim; i++) {x[i] = y[i] = 0;} }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%d", &n);int sum = 0;for (int i = 1; i <= n; i++) {scanf("%d", &A[i]);sum += A[i];}for (int i = 1; i <= n; i++) {scanf("%d", &B[i]);}init();solve(1, 1, n);for (int i = 0; i <= sum; i++) {printf("%d%c", f[1][i], i == sum ? '\n' : ' ');}return 0; }

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