无限手套
每种宝石的生成函数为∑n≥0xn(ain2+bin+1)对其进行化简∑n≥0xn+∑n≥0binxn+∑n≥0ain2xn11−x+bix(1−x)2+aix(1+x)(1−x)3最后答案∏i=1m((ai−bi+1)x2+(ai+bi−2)x+1)(1−x)3m每种宝石的生成函数为\sum_{n \geq 0} x ^ n(a_i n ^ 2 + b_i n + 1)\\ 对其进行化简\sum_{n \geq 0} x ^ n + \sum_{n \geq 0} b_i n x ^ n + \sum_{n\geq 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\limits_{i = 1} ^{m} \left((a_i - b_i + 1) x ^ 2 + (a_i + b_i - 2) x + 1 \right)}{(1 - x) ^{3m}}\\ 每种宝石的生成函数为n≥0∑xn(ain2+bin+1)对其进行化简n≥0∑xn+n≥0∑binxn+n≥0∑ain2xn1−x1+(1−x)2bix+(1−x)3aix(1+x)最后答案(1−x)3mi=1∏m((ai−bi+1)x2+(ai+bi−2)x+1)
#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;
}
