(mathcal{Description})
Link.
用 (m) 种颜色为长为 (n) 的序列染色，每个位置一种颜色。对于一种染色方案，其价值为 (w( ext{出现恰 }s ext{ 次的颜色种数}))（(w(0..m)) 给定），求所有染色方案的价值和。
(nle10^7)，(mle10^5)，答案对 (p=1004535809=479 imes2^{21}+1) 取模。
(mathcal{Solution})
记 (l=min{m,lfloorfrac{n}{s} floor})，显然只能对于 (i=0..l)，分别算出 (w(i)) 的贡献次数。考虑容斥，令 (f(i)) 表示至少 (i) 种颜色恰好出现 (s) 次的染色方案数。那么：
[f(i)=inom{m}ifrac{n!}{(s!)^i(n-is)!}(m-i)^{n-is}
]
即：选出 (i) 种颜色钦定恰好出现 (s) 次；多重集排列安排已被染色的 (is) 个位置和剩下的 ((n-is)) 个位置；为 ((n-is)) 个位置任选未使用的颜色。
此后，令 (g(i)) 表示恰好 (i) 种颜色恰好出现 (s) 次的染色方案数，容斥：
[egin{aligned}
g(i)&=sum_{j=i}^l(-1)^{l-i}inom{j}if_j\
&=sum_{j=i}^l(-1)^{l-j}frac{j!}{i!(j-i)!}f_j\
&=frac{1}{i!}sum_{j=i}^lfrac{(-1)^{l-j}}{(j-i)!}(j!f_j)
end{aligned}
]
很俗套 trick，翻转乘积式任意一项就能写成卷积形式，NTT 求出 (g(i)) 即可。
复杂度 (mathcal O(n+llog l))。
(mathcal{Code})
/* Clearink */
#include <cstdio>
#define rep( i, l, r ) for ( int i = l, rpbound##i = r; i <= rpbound##i; ++i )
#define per( i, r, l ) for ( int i = r, rpbound##i = l; i >= rpbound##i; --i )
const int MAXN = 1e7, MAXM = 1e5, MOD = 1004535809, G = 3, MAXLEN = 1 << 18;
int n, m, s, w[MAXM + 5], f[MAXLEN + 5], g[MAXLEN + 5], rev[MAXLEN + 5];
int fac[MAXN + 5], ifac[MAXN + 5]; // warning: 77MB.
inline char fgc () {
static char buf[1 << 17], *p = buf, *q = buf;
return p == q && ( q = buf + fread ( p = buf, 1, 1 << 17, stdin ), p == q )
? EOF : *p++;
}
inline int rint () {
int x = 0; char s = fgc ();
for ( ; s < '0' || '9' < s; s = fgc () );
for ( ; '0' <= s && s <= '9'; s = fgc () ) x = x * 10 + ( s ^ '0' );
return x;
}
inline void iswp ( int& a, int& b ) { a ^= b ^= a ^= b; }
inline int imin ( const int a, const int b ) { return a < b ? a : b; }
inline int imax ( const int a, const int b ) { return a < b ? b : a; }
inline int mul ( const long long a, const int b ) { return a * b % MOD; }
inline int sub ( int a, const int b ) { return ( a -= b ) < 0 ? a + MOD : a; }
inline int add ( int a, const int b ) { return ( a += b ) < MOD ? a : a - MOD; }
inline int mpow ( int a, int b ) {
int ret = 1;
for ( ; b; a = mul ( a, a ), b >>= 1 ) ret = mul ( ret, b & 1 ? a : 1 );
return ret;
}
inline int inv ( const int n ) { return mpow ( n, MOD - 2 ); }
inline void init ( const int n ) {
fac[0] = 1;
rep ( i, 1, n ) fac[i] = mul ( i, fac[i - 1] );
ifac[n] = inv ( fac[n] );
per ( i, n - 1, 0 ) ifac[i] = mul ( i + 1, ifac[i + 1] );
}
inline int comb ( const int n, const int m ) {
return n < m ? 0 : mul ( fac[n], mul ( ifac[m], ifac[n - m] ) );
}
inline void NTT ( const int n, int* A, const int flg ) {
rep ( i, 0, n - 1 ) if ( i < rev[i] ) iswp ( A[i], A[rev[i]] );
for ( int i = 2, stp = 1; i <= n; i <<= 1, stp <<= 1 ) {
int w = mpow ( G, ( MOD - 1 ) / i );
if ( !~flg ) w = inv ( w );
for ( int j = 0; j < n; j += i ) {
for ( int k = j, wk = 1; k < j + stp; ++k, wk = mul ( wk, w ) ) {
int ev = A[k], ov = mul ( wk, A[k + stp] );
A[k] = add ( ev, ov ), A[k + stp] = sub ( ev, ov );
}
}
}
if ( !~flg ) {
int in = inv ( n );
rep ( i, 0, n - 1 ) A[i] = mul ( in, A[i] );
}
}
int main () {
n = rint (), m = rint (), s = rint ();
init ( imax ( n, m ) );
rep ( i, 0, m ) w[i] = rint ();
int mx = imin ( m, n / s );
for ( int i = 0, fp = 1; i <= mx; ++i, fp = mul ( fp, fac[s] ) ) {
f[i] = mul ( fac[i], mul ( mul ( comb ( m, i ), mpow ( m - i, n - i * s ) ),
mul ( fac[n], inv ( mul ( fp, fac[n - i * s] ) ) ) ) );
}
rep ( i, 0, mx ) {
g[i] = ( ( mx - i ) & 1 ? sub : add )( 0, inv ( fac[mx - i] ) );
}
int len = 1, lgv = 0;
for ( ; len < mx + 1 << 1; len <<= 1, ++lgv );
rep ( i, 0, len - 1 ) rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << lgv >> 1 );
NTT ( len, f, 1 ), NTT ( len, g, 1 );
rep ( i, 0, len - 1 ) f[i] = mul ( f[i], g[i] );
NTT ( len, f, -1 );
int ans = 0;
rep ( i, 0, mx ) {
ans = add ( ans, mul ( w[i], mul ( f[i + mx], ifac[i] ) ) );
}
printf ( "%d
", ans );
return 0;
}