[HAOI2018]染色

嘟嘟嘟




这题当时没想出来（因为本人实在不太擅长计数），然后又被luogu的第一篇题解吓怕了，就咕了一小段时间再写。

其实这题不是很难。




做法就是基础容斥+NTT。




首先出现\(S\)次的颜色最多有\(N = min \{ \frac{n}{S}, m \}\)种。

我们令\(dp[i]\)表示出现\(S\)次的颜色至少有\(i\)种的方案数，那么共有\(C_{m} ^ {i}\)种颜色组合，这些颜色的位置共有\(C_{n} ^ {iS}\)种选取方案，剩下的位置每一位都有\(m\)中颜色可选，然后再考虑这\(C_{n} ^{iS}\)个位置中每一种颜色的分配方案，就有

\[\begin{align*}
dp[i]
&= C_{m} ^ {i} * C _{n} ^ {iS} * (C_{iS} ^ {S} * C_{iS - S} ^ {S} * C_{iS - 2S} ^ {S} * \ldots * C_{S} ^ {S}) * m ^ {n - iS} \\
&= C_{m} ^ {i} * C _{n} ^ {iS} * \frac{(iS)!}{(S!) ^ i} * m ^ {n - iS} \\
\end{align*}
\]

然后我们令\(ans[i]\)表示出现\(S\)次的颜色恰好有\(i\)种的方案数，根据容斥，就有这么个式子：

\[\begin{align*}
ans[i] &= \sum _ {j = i} ^ {N} (-1) ^ {j - i} C_{j} ^ {i} dp[j] \\
ans[i] *i! &= \sum _ {j = i} ^ {N} \frac{(-1) ^ {j - i}}{(j - i)!} * dp[j] * j!
\end{align*}
\]

这个东西NTT可做。把dp数组反过来。就像[ZJOI2014]力这道题一样。

需要注意的是这样\(ans[i]\)也是反过来的，所以乘上的是\(inv[N - i]\)和\(w[N - i]\)。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 2e7 + 5;
const ll mod = 1004535809;
const ll G = 3;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, m, S, N, w[maxn];

In ll inc(ll a, ll b) {return a + b >= mod ? a + b - mod : a + b;}
In ll quickpow(ll a, ll b)
{
  ll ret = 1;
  for(; b; b >>= 1, a = a * a % mod)
    if(b & 1) ret = ret * a % mod;
  return ret;
}

ll fac[maxn], inv[maxn];
In void init()
{
  int Max = max(n, m);
  fac[0] = inv[0] = 1;
  for(int i = 1; i <= Max; ++i) fac[i] = fac[i - 1] * i % mod;
  inv[Max] = quickpow(fac[Max], mod - 2);
  for(int i = Max - 1; i; --i) inv[i] = inv[i + 1] * (i + 1) % mod;
}

ll dp[maxn], b[maxn];
int len = 1, lim = 0, rev[maxn];
In void ntt(ll* a, int len, int flg)
{
  for(int i = 0; i < len; ++i) if(i < rev[i]) swap(a[i], a[rev[i]]);
  for(int i = 1; i < len; i <<= 1)
    {
      ll gn = quickpow(G, (mod - 1) / (i << 1));
      for(int j = 0; j < len; j += (i << 1))
	  {
	    ll g = 1;
	    for(int k = 0; k < i; ++k, g = g * gn % mod)
	    {
	      ll tp1 = a[k + j] % mod, tp2 = g * a[k + j + i] % mod;
	      a[k + j] = inc(tp1, tp2), a[k + j + i] = inc(tp1, mod - tp2);
	      //a[k + j] = (tp1 + tp2) % mod, a[k + j + i] = (tp1 - tp2 + mod) % mod;
	    }
	  }
    }
  if(flg == 1) return;
  ll inv = quickpow(len, mod - 2); reverse(a + 1, a + len);
  for(int i = 0; i < len; ++i) a[i] = a[i] * inv % mod;
}

int Ans[maxn];
In void bf()
{
  for(int i = 0; i <= N; ++i)
    for(int j = 0; j <= i; ++j)
      Ans[i] = inc(Ans[i], dp[j] * b[i - j] % mod);
  ll ans = 0;
  for(int i = 0; i <= N; ++i) ans = inc(ans, Ans[i] * inv[N - i] % mod * w[N - i] % mod);
  printf("--->%lld\n", ans);
}

int main()
{
  n = read(), m = read(), S = read();
  N = min(m, n / S);
  for(int i = 0; i <= m; ++i) w[i] = read();
  init();
  for(int i = 0; i <= N; ++i)
    dp[i] = fac[n] * fac[m] % mod * inv[i] % mod * inv[m - i] % mod * quickpow(inv[S], i) % mod * inv[n - i * S] % mod * quickpow(m - i, n - i * S) % mod * fac[i] % mod;
  for(int i = 0; i <= N; ++i) b[i] = (i & 1) ? mod - inv[i] : inv[i];
  reverse(dp, dp + N + 1);
  //bf();
  while(len <= (N << 1)) len <<= 1, ++lim;
  for(int i = 0; i < len; ++i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (lim - 1));
  ntt(dp, len, 1), ntt(b, len, 1);
  for(int i = 0; i < len; ++i) dp[i] = dp[i] * b[i] % mod;
  ntt(dp, len, -1);
  ll ans = 0;
  for(int i = 0; i <= N; ++i) ans = inc(ans, dp[i] * inv[N - i] % mod * w[N - i] % mod);
  write(ans), enter;
  return 0;
}