有一块长度为 nn 的画布,每个位置可以染成 [1,m][1, m] 这些颜色中的一种。

如果画布上恰好有 kk 种颜色恰好出现了 ss 次,则会产生 wkw_k 的愉悦度,求所有不同画布的愉悦度之和,对 10045358091004535809 取模。

n≤106,m≤106,s≤150n\le 10^6, m\le 10^6, s\le 150

LOJ 2527

Solution

感觉是一道容斥的入门题。。。

设fif_i表示钦定ii种颜色出现了恰好ss次的方案数，gig_i表示恰好ii种颜色出现了ss次的方案数，lim=min{m,ns}lim = \min\{m, \frac{n}{s}\}

fi=(mi)(nis)(is)!(s!)i×(m−i)n−is \begin{aligned} f_i &= \binom{m}{i}\binom{n}{is}\frac{(is)!}{(s!)^i}\times(m-i)^{n-is}\\ \end{aligned} gi=∑j=ilim(−1)j−i(ji)fj=1i!∑j=ilim(−1)j−i(j−i)!×j!⋅fj \begin{aligned} g_i &= \sum_{j=i}^{lim}(-1)^{j-i}\binom{j}{i}f_j\\ &=\frac{1}{i!}\sum_{j=i}^{lim}\frac{(-1)^{j-i}}{(j-i)!}\times j!\cdot f_j\\ \end{aligned}

直接卷积即可

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> inline T read ()
{
	T sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

inline void proc_status ()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

const int Maxn = 2e6 + 100;
const int Mod = 1004535809;

int N, M, S;
int W[Maxn];

namespace MATH
{
	const int maxn = 1e7 + 5;
	int fac[maxn], ifac[maxn];

	inline void Add (int &a, int b) { if ((a += b) >= Mod) a -= Mod; }

	inline int Pow (int a, int b)
	{
		int ans = 1;
		for (int i = b; i; i >>= 1, a = (LL) a * a % Mod) if (i & 1) ans = (LL) ans * a % Mod;
		return ans;
	}

	inline void init (int n)
	{
		fac[0] = 1;
		for (int i = 1; i <= n; ++i) fac[i] = (LL) fac[i - 1] * i % Mod;
		ifac[n] = Pow (fac[n], Mod - 2);
		for (int i = n - 1; i >= 0; --i) ifac[i] = (LL) ifac[i + 1] * (i + 1) % Mod;
	}

	inline int C (int n, int m) { if (n < m) return 0; return (LL) fac[n] * ifac[m] % Mod * ifac[n - m] % Mod; }
}

using namespace MATH;

namespace Poly
{
	const int g = 3;
	int n, rev[Maxn];

	inline void dft (int *A, int fg)
	{
		for (int i = 0; i < n; ++i) if (i < rev[i]) swap (A[i], A[rev[i]]);
		for (int mid = 1; mid < n; mid <<= 1)
		{
			int Wn = Pow (g, (Mod - 1) / (mid << 1));
			if (fg == -1) Wn = Pow (Wn, Mod - 2);
			for (int i = 0; i < n; i += mid << 1)
				for (int j = i, W = 1; j < i + mid; ++j, W = (LL) W * Wn % Mod)
				{
					int x = A[j], y = (LL) W * A[j + mid] % Mod;
					A[j] = (x + y) % Mod;
					A[j + mid] = (x - y + Mod) % Mod;
				}
		}
		if (fg == -1) for (int i = 0, inv = Pow (n, Mod - 2); i < n; ++i) A[i] = (LL) A[i] * inv % Mod;
	}

	inline void mul (int *A, int *B, int *C, int N)
	{
		for (n = 1; n <= N << 1; n <<= 1);
		for (int i = 0; i < n; ++i) rev[i] = (rev[i >> 1] >> 1) + ((i & 1) ? (n >> 1) : 0);

		static int F[Maxn], G[Maxn];

		for (int i = 0; i < n; ++i) F[i] = (i <= N) ? A[i] : 0;
		for (int i = 0; i < n; ++i) G[i] = (i <= N) ? B[i] : 0;

		dft (F, 1), dft (G, 1);
		for (int i = 0; i < n; ++i) F[i] = (LL) F[i] * G[i] % Mod;
		dft (F, -1);

		for (int i = 0; i <= N << 1; ++i) C[i] = F[i];
	}
}

int F[Maxn], G[Maxn], H[Maxn];

inline void Solve ()
{
	int LIM = min (M, N / S);
	for (int i = 0; i <= LIM; ++i) F[i] = (LL) C (M, i) * C (N, i * S) % Mod * fac[i * S] % Mod * Pow (ifac[S], i) % Mod * Pow (M - i, N - i * S) % Mod * fac[i] % Mod;
	for (int i = 0; i <= LIM; ++i) G[i] = (LL) Pow (Mod - 1, i) * ifac[i] % Mod;
	reverse (F, F + LIM + 1);

	Poly :: mul (F, G, H, LIM);

	reverse (H, H + LIM + 1);

	int ans = 0;
	for (int i = 0; i <= LIM; ++i) Add (ans, (LL) H[i] * ifac[i] % Mod * W[i] % Mod);
	cout << ans << endl;
}

inline void Input ()
{
	N = read<int>(), M = read<int>(), S = read<int>();
	for (int i = 0; i <= M; ++i) W[i] = read<int>();
}

int main()
{

#ifdef hk_cnyali
	freopen("A.in", "r", stdin);
	freopen("A.out", "w", stdout);
#endif

	MATH :: init (1e7);
	Input ();
	Solve ();

	return 0;
}