無限手套

$$\begin{gathered} 每種寶石的生成函數為\sum _{n\ge 0}{x}^n(a_i{n}^2+b_{i}n+1) \\ 對其進行化簡\sum _{n\ge 0}{x}^n+\sum _{n\ge 0}{b}_{i}n{x}^n+\sum _{n\ge 0}{a}_i{n}^2{x}^n \\ \frac{1}{1?x}+\frac{b_{i}x}{(1?x{)}^2}+\frac{a_{i}x(1+x)}{(1?x{)}^3} \\ 最后答案\frac{\prod _{i=1}^m\left((a_i?b_i+1)x^2+(a_i+b_i?2)x+1\right)}{(1?x{)}^{3m}} \end{gathered}$$

#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10, mod = 998244353;int fac[N], inv[N];int quick_pow(int a, int n) {int ans = 1;while (n) {if (n & 1) {ans = 1ll * ans * a % mod;}a = 1ll * a * a % mod;n >>= 1;}return ans; }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] = 1;for(int i = 1; i < N; i++)fac[i] = 1ll * fac[i - 1] * i % 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], n, m;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();scanf("%d", &m);A[0] = 1;for (int i = 1, a, b; i <= m; i++) {scanf("%d %d", &a, &b);int x = (a - b + 1 + mod) % mod, y = (a + b - 2) % mod, z = 1;for (int i = 2 * m; i >= 2; i--) {A[i] = (1ll * A[i - 2] * x % mod + 1ll * A[i - 1] * y % mod + 1ll * A[i] * z % mod) % mod;}A[1] = (1ll * A[0] * y % mod + 1ll * A[1] * z % mod) % mod;A[0] = 1ll * A[0] * z % mod;}int T;scanf("%d", &T);while (T--) {scanf("%d", &n);int ans = 0;for (int i = 0; i <= n; ++i) {ans = (ans + 1ll * C(3 * m + i - 1, i) * A[n - i] % mod) % mod;}printf("%d\n", ans);}return 0; }

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