P4491 [HAOI2018]染色

P4491 [HAOI2018]染色

考虑每个 $K$ 的贡献，在 $m$ 个颜色里选择 $K$ 个，有 $\binom m K$ 种方案。选择的这些颜色必须出现 $S$ 次，那么从 $n$ 个位置中选择 $n - SK$ 个位置中，每个位置都有 $m - K$ 种选择，即 $\binom n {n - SK} (m - K) ^ {n - SK}$ 。对于这 $SK$ 个位置中，选择的颜色也是固定的了，只需要考虑排列顺序，而每种颜色中顺序是无所谓的，即 $\frac {(SK)!} {(S!) ^ K}$ 。注意到这样计算后其他位置的颜色是不可控制的，答案是算多了的。记这个式子 $F(K) = \binom m K \binom n {n - SK} (m - K) ^ {n - SK} \frac {(SK)!} {(S!) ^ K} = \frac {m!n!(m - K) ^ {n - SK}}{K!(m - K)!(n - SK)!(S!) ^ K}$ 。

考虑其和真实答案的关系，枚举多少个多的被算上了，即 $F(x) = \sum _ {i= x} \binom m i G(i)$ 。二项式反演，那么有 $G(x) = \sum _ {i = x} (-1) ^ {i - x} \binom i x \frac {m!n!(m - i) ^ {n - Si}}{i!(m - i)!(n - Si)!(S!) ^ K} $ 。

将关于 $i$ 和 $i - x$ 的项提出，就可以卷积计算了。

查看代码

#include <cstdio>
#include <algorithm>
using namespace std;
template <class Type>
void read (Type &x)
{
    char c;
    bool flag = false;
    while ((c = getchar()) < '0' || c > '9')
        flag |= c == '-';
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 3) + (x << 1) + c - '0';
    if (flag) x = ~x + 1;
}
template <class Type, class ...Rest>
void read (Type &x, Rest &...y) { read(x); read(y...); }
template <class Type>
void write (Type x)
{
    if (x < 0) putchar('-'), x = ~x + 1;
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
}
typedef long long LL;
const int N = 1e7  + 10, mod = 1004535809;
int rev[N];
int n, m, S;
int inv[N], fact[N], ifact[N];
int binpow (int b, int k = mod - 2)
{
    int res = 1;
    for (; k; k >>= 1, b = (LL)b * b % mod)
        if (k & 1) res = (LL)res * b % mod;
    return res;
}
void init ()
{
    inv[1] = 1;
    for (int i = 2; i < N; ++i)
        inv[i] = -(LL)(mod / i) * inv[mod % i] % mod;
    fact[0] = ifact[0] = 1;
    for (int i = 1; i < N; ++i)
    {
        fact[i] = (LL)fact[i - 1] * i % mod;
        ifact[i] = (LL)ifact[i - 1] * inv[i] % mod;
    }
}
int C (int a, int b) { return a < b ? 0 : (LL)fact[a] * ifact[b] % mod * ifact[a - b] % mod; }
void ntt (int *x, int bit, int op)
{
    int tot = 1 << bit;
    for (int i = 0; i < tot; ++i)
        if ((rev[i] = rev[i >> 1] >> 1 | (i & 1) << bit - 1) > i)
            swap(x[i], x[rev[i]]);
    for (int mid = 1; mid < tot; mid <<= 1)
    {
        int w1 = binpow(3, (mod - 1) / (mid << 1));
        if (!~op) w1 = binpow(w1);
        for (int i = 0; i < tot; i += mid << 1)
            for (int j = 0, k = 1; j < mid; ++j, k = (LL)k * w1 % mod)
            {
                int p = x[i + j], q = (LL)k * x[i + j + mid] % mod;
                x[i + j] = (p + q) % mod, x[i + j + mid] = (p - q) % mod;
            }
    }
    if (~op) return;
    int itot = binpow(tot);
    for (int i = 0; i < tot; ++i)
        x[i] = (LL)x[i] * itot % mod;
}
int main ()
{
    init();
    read(n, m, S);
    int len = min(m, n / S);
    static int A[N], B[N];
    int bit = 0;
    while (len + len + 1 > 1 << bit) ++bit;
    int tot = 1 << bit;
    for (int i = 0; i <= len; ++i)
    {
        A[i] = (LL)fact[i] * C(m, i) % mod * binpow(ifact[S], i) % mod * ifact[n - S * i] % mod * binpow(m - i, n - S * i) % mod;
        B[len - i] = (i & 1 ? -1 : 1) * ifact[i];
    }
    for (int i = len + 1; i < tot; ++i)
        A[i] = B[i] = 0;
    ntt(A, bit, 1), ntt(B, bit, 1);
    for (int i = 0; i < tot; ++i)
        A[i] = (LL)A[i] * B[i] % mod;
    ntt(A, bit, -1);
    int res = 0;
    for (int i = 0, a; i <= len; ++i)
    {
        read(a);
        (res += (LL)a * ifact[i] % mod * A[len + i] % mod) %= mod;
    }
    write(((LL)res * fact[n] % mod + mod) % mod);
    return 0;
}