bzoj 5306 [HAOI2018] 染色

链接

推式子题

首先枚举有几种颜色选择恰好 (s) 次，可以得到一个式子：

[sum _{i = 0} ^ {min(frac n s,m )} frac {binom n {i cdot s} binom m i (i cdot s) ! cdot f_i cdot (m-i)^{n - is}} {(s!) ^ i} ]

但是，(f_i) 不能单纯地等于 (w_i),因为会重复计算，我们不能保证当前选择的 (i) 种颜色之外的颜色是否没有出现恰好 (s) 次的

发现对于一种染色方案而言，假设其中选择恰好 (s) 次的颜色数量为 (i)，那么他在枚举到 (j) 种颜色出现 (s) 次是被算进去 (binom i j) 次，所有被算进去的 (f) 值之和就是 (sum _{j = 0} ^ i (f_j cdot binom i j))

我们令 (w_i = sum _{j = 0} ^i (f_j cdot binom i j))， 那么这样可以推出来 (f) 序列，并且把这里的 (f) 序列代入之前的式子就可以求出答案了

怎么求 (f) 呢？有两种方式

1. 发现 (sum _{j = 0} ^ i frac{f_j} {j !} cdot frac 1 {(i - j) !} = frac {w_i} {i!})，可以用多项式除法直接求得 (frac {f_j} {j!})

2. 倒推。

   $w_0 = f_0 $

   (w_1 = f_0 + f_1)

   (w_2=f_0+2f_1+f_2)

   (w_3=f_0+3f_1+3f_2+f_3)

   (dots)

   解得

   (f_0=w_0)

   (f_1=w_1-w_0)

   (f_2=w_2-2w_1+w_0)

   (f_3=w_3-3w_2+3w_1-w_0)

   找规律，就是 (f_i = sum _{j = 0} ^ i (-1)^{i-j} cdot binom i j cdot w_j = i! cdot sum_{j=0}^i frac {(-1)^{i-j}} {(i-j)!} cdot frac {w_j} {j!})

   卷积即可求出 (f_i)

惊讶地发现，(1004535809) 这个模数的原根也是 (3)

代码

// Copyright lzt
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<iostream>
#include<queue>
#include<string>
#include<ctime>
using namespace std;
typedef long long ll;
typedef std::pair<int,int> pii;
typedef long double ld;
typedef unsigned long long ull;
typedef std::pair<long long,long long> pll;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define rep(i,j,k)  for (register int i = (int)(j); i <= (int)(k); i++)
#define rrep(i,k) for (register int i = (int)(j); i >= (int)(k); i--)
#define Debug(...) fprintf(stderr,__VA_ARGS__)

inline ll read() {
  ll x = 0,f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * f;
}

const int mod = 1004535809;
const int GEN = 3;

namespace poly {
  int alpha[2][400400],rev[400400],ntt_lst = -1;
  inline int sum(int x,int y) {
    x += y;
    return x >= mod ? x - mod : x;
  }
  inline int sub(int x,int y) {
    x -= y;
    return x < 0 ? x + mod : x;
  }
  inline int ksm(int x,int p) {
    int ret = 1;
    while (p) {
      if (p & 1) ret = ret * 1ll * x % mod;
      x = x * 1ll * x % mod; p >>= 1;
    }
    return ret;
  }
  inline void ntt_init(int n) {
    if (ntt_lst == n) return;
    ntt_lst = n;
    alpha[0][0] = alpha[1][0] = 1;
    alpha[0][1] = ksm(GEN,(mod - 1) / n);
    alpha[1][1] = ksm(alpha[0][1],mod - 2);
    rep(i,2,n) rep(x,1) alpha[x][i] = alpha[x][i - 1] * 1ll * alpha[x][1] % mod;
    int nw = n >> 1;
    rep(i,1,n - 2) {
      rev[i] = nw;
      int j = n >> 1;
      while (nw >= j) {
        nw -= j;
        j >>= 1;
      }
      nw += j;
    }
  }
  inline void ntt(int *a,int n,bool f) {
    ntt_init(n);
    rep(i,n - 2) if (i < rev[i]) swap(a[i],a[rev[i]]);
    for (int i = 1; i < n; i <<= 1) {
      for (int j = 0,off = n / (i << 1); j + i < n; j += (i << 1)) {
        for (int k = j,cur = 0; k < j + i; k++,cur += off) {
          int x = a[k],y = a[k + i] * 1ll * alpha[f][cur] % mod;
          a[k] = sum(x,y); a[k + i] = sub(x,y);
        }
      }
    }
    if (f) {
      int x = ksm(n,mod - 2);
      rep(i,n - 1) a[i] = a[i] * 1ll * x % mod;
    }
  }
  inline void mul(int *a,int *b,int n) {
    ntt(a,n,false);
    ntt(b,false);
    rep(i,n - 1) a[i] = a[i] * 1ll * b[i] % mod;
    ntt(a,true);
  }
  inline void mul(int *a,int m,int *res,int mx_len = -1) {
    static int A[400400],B[400400];
    if (mx_len == -1) mx_len = n + m;
    n = min(n,mx_len); m = min(m,mx_len);
    int len = 1;
    while (len < n + m) len <<= 1;
    rep(i,len - 1) {
      A[i] = i < n ? a[i] : 0;
      B[i] = i < m ? b[i] : 0;
    }
    mul(A,B,len);
    memcpy(res,A,mx_len << 2);
  }
}

const int maxn = 400400;
int n,m,s;
int w[maxn],a[maxn],b[maxn],f[maxn];
int fac[10000100],inv[10000100];

inline int C(int x,int y) {
  return fac[x] * 1ll * inv[y] % mod * inv[x - y] % mod;
}

void work() {
  n = read(),m = read(),s = read();
  rep(i,m) w[i] = read();
  int lim = min(n / s,m),up = max(n,m);
  fac[0] = 1;
  rep(i,up) fac[i] = fac[i - 1] * 1ll * i % mod;
  inv[up] = poly::ksm(fac[up],mod - 2);
  rrep(i,up - 1,0) inv[i] = inv[i + 1] * 1ll * (i + 1) % mod;
  rep(i,lim) {
    a[i] = w[i] * 1ll * inv[i] % mod;
    if (i & 1) b[i] = (mod - inv[i]) % mod;
    else b[i] = inv[i] % mod;
  }
  poly::mul(a,lim + 1,b,a);
  rep(i,lim) f[i] = fac[i] * 1ll * a[i] % mod;
  int ans = 0;
  rep(i,lim) {
    int nw = C(n,i * s) * 1ll * C(m,i) % mod * fac[i * s] % mod * poly::ksm(inv[s],i) % mod * f[i] % mod * poly::ksm(m - i,n - i * s) % mod;
    ans = (ans + nw) % mod;
  }
  printf("%dn",ans);
}

int main() {
  #ifdef LZT
    freopen("in","r",stdin);
  #endif

  work();

  #ifdef LZT
    Debug("My Time: %.3lfmsn",(double)clock() / CLOCKS_PER_SEC);
  #endif
}