2022牛客暑假多校第四场C.Easy Counting Problem
作者：互联网
感谢 Pedestrian1 的指导
题意：\(q\) 次询问，每次问在一定约束条件下构造出长度为 \(n\) 的序列的方案数，只能用 \(w\) 种数字构造，且每种数字至少使用 \(c_i\) 次
知识点：多项式，生成函数
首先知道这个是要使用EGF的相关生成函数解决，然后我们先推式子
设对于数字 \(i\) 的方案生成函数为 \(A_i(x)\) 那么设答案多项式为 \(F(x)\)
则有 \(F(x) = \prod_{i=0}^{w-1} A_i(x)\)
其中 \(A_i(x) = \sum_{j=c_i} \frac{x^i}{i!}\)
即 \(F(x) = \prod_{i=0}^{w-1} (e^x - \sum_{i=0}^{c_i-1}\frac{x^i}{i!})\)
那么对于每次询问给出的 \(n\)，答案就是 \([x^n]F(x)\) ，现在问题就是如何求这个
显然，如果我们每次询问都是这么去直接卷，肯定是会超时的，那么我们考虑其他方法
考虑到 \(w\) 与后面的 \(\sum c_i\) 的范围很小，于是我们考虑展开这个式子
其中我们定义 \(f_i(x) = \sum_{i=0}^{c_i-1}\frac{x^i}{i!}\)
\(g_k(x)\) 表示 \(w\) 个 \(f\) 中选 \(k\) 个相乘的和，是用多项式来表达的（实际意义就是 \(w\) 个数任意选 \(k\) 个来组的方案数）
\(F(x) = \sum_{i=0}^{w} (-1)^i*e^{(w-i)x}*g_i(x)\)
注意到，如果我们想求 \([x^n]F(x)\) 我们可以对于每个 \(i\) 求出 \(\sum_{j=0}^n[x^j]e^{(w-i)x}*[x^{n-j}]g_i(x)\) ，然后求和即可
\(g\) 部分可以用 \(2^w\) 枚举或者 \(w^2\) 的背包求出，这里用背包写
这题感觉又卡时间又卡空间的，注意换一个比较好的板子。
#include <bits/stdc++.h>
#define endl '\n'
#define ls u << 1
#define rs u << 1 | 1
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
const int INF = 0x3f3f3f3f, N = 1e7 + 10;
const int MOD = 998244353;
const double eps = 1e-6;
const double PI = acos(-1);
inline int lowbit(int x) {return x & (-x);}
namespace polybase {//范围为1e9需要先取模,别忘记改模数,原根和数组长度
constexpr ll mod = 998244353, g = 3, L = 1 << 22;
ll fpow(ll x, ll r)
{
ll result = 1;
while (r)
{
if (r & 1)result = result * x % mod;
r >>= 1;
x = x * x % mod;
}
return result;
}
int w[L], _ = []
{
ll x = fpow(g, (mod - 1) / L);
w[L / 2] = 1;
for (int i = L / 2 + 1; i < L; i++)
w[i] = w[i - 1] * x % mod;
for (int i = L / 2 - 1; i >= 0; i--)
w[i] = w[i << 1];
return 0;
}();
inline int norm(int n) { return 1 << __lg(n * 2 - 1); }
void dft(ll *a, int n)
{
for (int k = n >> 1; k; k >>= 1)
for (int i = 0; i < n; i += k << 1)
for (int j = 0; j < k; j++)
{
ll &x = a[i + j], y = a[i + j + k];
a[i + j + k] = (x - y + mod) * w[k + j] % mod;
x += y;
if (x >= mod)x -= mod;
}
}
void idft(ll *a, int n)
{
for (int k = 1; k < n; k <<= 1)
for (int i = 0; i < n; i += k << 1)
for (int j = 0; j < k; j++)
{
ll x = a[i + j], y = a[i + j + k] * w[k + j] % mod;
a[i + j + k] = x - y < 0 ? x - y + mod : x - y;
a[i + j] += y;
if (a[i + j] >= mod)a[i + j] -= mod;
}
for (int i = 0, inv = mod - (mod - 1) / n; i < n; i++)
a[i] = a[i] * inv % mod;
reverse(a + 1, a + n);
}
struct poly : public vector<ll>
{
using vector<ll>::vector;
#define T (*this)
poly modxk(int k) const
{
k = min(k, (int) size());
return poly(begin(), begin() + k);
}
poly rev() const { return poly(rbegin(), rend()); }
friend void dft(poly &a) { dft(a.data(), a.size()); }
friend void idft(poly &a) { idft(a.data(), a.size()); }
friend poly operator*(const poly &x, const poly &y)
{
if (x.empty() || y.empty())return poly();
poly a(x), b(y);
int len = a.size() + b.size() - 1;
int n = norm(len);
a.resize(n), b.resize(n);
dft(a), dft(b);
for (int i = 0; i < n; i++)
a[i] = a[i] * b[i] % mod;
idft(a);
a.resize(len);
return a;
}
poly operator+(const poly &b)
{
poly a(T);
if (a.size() < b.size())
a.resize(b.size());
for (int i = 0; i < b.size(); i++)//用b.size()防止越界
{
a[i] += b[i];
if (a[i] >= mod)a[i] -= mod;
}
return a;
}
poly operator-(const poly &b)
{
poly a(T);
if (a.size() < b.size())
a.resize(b.size());
for (int i = 0; i < b.size(); i++)
{
a[i] -= b[i];
if (a[i] < 0)a[i] += mod;
}
return a;
}
poly operator*(const ll p)
{
poly a(T);
for (auto &x:a)
x = x * p % mod;
return a;
}
poly &operator<<=(int r) { return insert(begin(), r, 0), T; }//注意逗号,F(x)*(x^r)
poly operator<<(int r) const { return poly(T) <<= r; }
poly operator>>(int r) const { return r >= size() ? poly() : poly(begin() + r, end()); }
poly &operator>>=(int r) { return T = T >> r; }//F[x]/(x^r)
poly deriv()//求导
{
if (empty())return T;
poly a(size() - 1);
for (int i = 1; i < size(); i++)//注意是size()
a[i - 1] = T[i] * i % mod;
return a;
}
poly integ()//积分
{
poly a(size() + 1);
for (int i = 1; i < a.size(); i++)//注意是a.size()
a[i] = T[i - 1] * fpow(i, mod - 2) % mod;
return a;
}
poly inv(int n)
{
poly a{fpow(T[0], mod - 2)};
int k = 1;
while (k < n)
{
k <<= 1;
a = (a * 2 - modxk(k) * a * a).modxk(k);
}
return a.modxk(n);
}
poly sqrt(int n)//f[0]必须等于1
{
poly a{1};
int k = 1;
const ll inv2 = fpow(2, mod - 2);
while (k < n)
{
k <<= 1;
a = ((modxk(k) * a.inv(k)).modxk(k) + a) * inv2;
}
return a.modxk(n);
}
poly ln(int n)//需要保证f[0]=1
{
return (deriv() * inv(n)).integ().modxk(n);
}
poly exp(int n)//需要保证f[0]=0
{
poly a{1};
int k = 1;
while (k < n)
{
k <<= 1;
a = (a * (poly{1} - a.ln(k) + modxk(k))).modxk(k);
}
return a.modxk(n);
}
#undef T
};
}
using namespace polybase;
int fac[N], ifac[N];
poly G[11], F[11];
inline void solve() {
int w; cin >> w;
for (int i = 0; i < w; i ++ ) {
int x; cin >> x;
G[i].resize(x);
for (int j = 0; j < x; j ++ ) G[i][j] = ifac[j];
}
F[0] = poly{1};
for (int i = 0; i < w; i ++ ) for (int j = i + 1; j; j -- ) F[j] = F[j] + F[j - 1] * G[i];
int m; cin >> m;
while (m -- ) {
int n; cin >> n;
ll res = 0;
for (int i = 0; i <= w; i ++ ) {
ll now = fpow(w - i, n);
ll nv = fpow(w - i, MOD - 2);
for (int j = 0; j < F[i].size() && j <= n; j ++ ) {
if (i == w && n == j) now = 1;
res = (res + 1ll * (i & 1 ? MOD - 1 : 1) * ifac[n - j] % MOD * now % MOD * F[i][j] % MOD) % MOD;
now = now * nv % MOD;
}
}
res = res * fac[n] % MOD;
cout << res << endl;
}
}
signed main() {
#ifdef DEBUG
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
auto now = clock();
#endif
ios::sync_with_stdio(false), cin.tie(nullptr);
cout << fixed << setprecision(2);
int n = 10000000;
fac[0] = 1;
for (int i = 1; i <= n; i ++ ) fac[i] = 1ll * fac[i - 1] * i % MOD;
ifac[n] = fpow(fac[n], MOD - 2);
for (int i = n; i; i -- ) ifac[i - 1] = 1ll * ifac[i] * i % MOD;
// int T; cin >> T;
// while (T -- )
solve();
#ifdef DEBUG
cout << "============================" << endl;
cout << "Program run for " << (clock() - now) / (double)CLOCKS_PER_SEC * 1000 << " ms." << endl;
#endif
return 0;
}
标签：第四场,int,多校,poly,牛客,const,return,mod,size 来源： https://www.cnblogs.com/JYF-AC/p/16538667.html