【LOJ】#2527. 「HAOI2018」染色

题解

简单容斥题

至少选了\(k\)个颜色恰好出现\(S\)次方案数是

\(F[k] = \binom{M}{k} \frac{N!}{(S!)^{k}(N - i * S)!}(M - k)^{N - i * S}\)

然后恰好\(k\)个颜色恰好出现\(k\)次就是

\(g[k] = \sum_{j = k}^{M} (-1)^{k - j} \binom{j}{k}F[j]\)

然后就是

\(g[k]*k! = \sum_{j = k}^{M}\frac{(-1)^{j - k}}{(j - k)!} F[j]*j!\)

后面的只要把其中一个指数取反，就可以NTT卷积优化了

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define MAXN 100005
#define mo 974711
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1004535809;
const int MAXL = 1 << 20;
int W[MAXL + 5],a[MAXN],N,M,S,F[MAXN];
int fac[10000005],invfac[10000005];
int f[1000005],g[1000005];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int C(int n,int m) {
    if(n < m) return 0;
    return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
	if(c & 1) res = mul(res,t);
	t = mul(t,t);
	c >>= 1;
    }
    return res;
}
void NTT(int *p,int L,int on) {
    for(int i = 1 , j = L >> 1 ; i < L - 1 ; ++i) {
	if(i < j) swap(p[i],p[j]);
	int k = L >> 1;
	while(j >= k) {
	    j -= k;
	    k >>= 1;
	}
	j += k;
    }
    for(int h = 2 ; h <= L ; h <<= 1) {
	int wn = W[(MAXL + on * MAXL / h) % MAXL];
	for(int k = 0 ; k < L ; k += h) {
	    int w = 1;
	    for(int j = k ; j < k + h / 2 ; ++j) {
		int u = p[j],t = mul(w,p[j + h / 2]);
		p[j] = inc(u,t);
		p[j + h / 2] = inc(u,MOD - t);
		w = mul(w,wn);
	    }
	}
    }
    if(on == -1) {
	int InvL = fpow(L,MOD - 2);
	for(int i = 0 ; i < L ; ++i) p[i] = mul(InvL,p[i]);
    }
}
void Solve() {
    read(N);read(M);read(S);
    for(int i = 0 ; i <= M ; ++i) read(a[i]);
    fac[0] = 1;
    for(int i = 1 ; i <= 10000000 ; ++i) fac[i] = mul(fac[i - 1],i);
    invfac[10000000] = fpow(fac[10000000],MOD - 2);
    for(int i = 10000000 - 1 ; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);
    W[0] = 1;
    W[1] = fpow(3,(MOD - 1) / MAXL);
    for(int i = 2 ; i < MAXL ; ++i) W[i] = mul(W[i - 1],W[1]);
    int t = 1;
    for(int i = 0 ; i <= M ; ++i) {
	if(N / S < i) break;
	F[i] = mul(C(M,i) , mul(fac[N] , mul(t , invfac[N - i * S])));
	F[i] = mul(F[i],fpow(M - i,N - i * S));
	t = mul(t,invfac[S]);
    }
    for(int i = 0 ; i <= M ; ++i) f[i] = mul(F[i],fac[i]);
    t = 1;
    for(int i = 0 ; i <= M ; ++i) {
	g[M - i] = mul(t,invfac[i]);
	t = mul(t,MOD - 1);
    }
    t = 1;
    while(t <= 2 * M) t <<= 1;
    NTT(f,t,1);NTT(g,t,1);
    for(int i = 0 ; i < t ; ++i) g[i] = mul(g[i],f[i]);
    NTT(g,t,-1);

    int ans = 0;
    for(int i = 0 ; i <= M ; ++i) {
	g[i + M] = mul(g[i + M],invfac[i]);
	ans = inc(ans,mul(g[i + M],a[i]));
    }

    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}