【LG4491】[HAOI2018]染色
题面
题解
颜色的数量不超过(lim=min(m,frac nS))
考虑容斥，计算恰好出现(S)次的颜色至少(i)种的方案数(f[i])，钦定(i)种颜色至少放(S)种
有(m)种颜色，那么要乘上(C_m^i)。
然后这(n)个位置分为(i+1)个部分：被钦定的(i)种颜色，每个(S)个；剩下(m-i)种颜色，一共(n-iS)种颜色，可以看作可重的全排列数，那么就有(frac{n!}{(S!)^i(n-iS)!})，但是后面情况的每个部分有(m-i)种取法，所以还要乘上((m-i)^{n-iS})
( herefore f[i]=C_m^ifrac{n!}{(S!)^i(n-iS)!}(m-i)^{n-iS})
那么答案是什么呢？
(ans[i]:)恰好出现(S)次的颜色恰好(i)种的方案数
用容斥搞一下：
[ans[i]=sum_{j=i}^{lim}(-1)^{j-i}C_j^if[j]\
=sum_{j=i}^{lim}(-1)^{j-i}frac{j!}{i!(j-i)!}f[j]\
Leftrightarrow
ans[i]*i!=sum_{j=i}^{lim}frac{(-1)^{j-i}}{(j-i)!}f[j]*j!
]
然后(NTT)就可以了。
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
inline int gi() {
register int data = 0, w = 1;
register char ch = 0;
while (!isdigit(ch) && ch != '-') ch = getchar();
if (ch == '-') w = -1, ch = getchar();
while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
return w * data;
}
const int Mod = 1004535809;
int fpow(int x, int y) {
int res = 1;
while (y) {
if (y & 1) res = 1ll * res * x % Mod;
x = 1ll * x * x % Mod;
y >>= 1;
}
return res;
}
const int G = 3, iG = fpow(G, Mod - 2);
const int MAX_N = 1e7 + 5, MAX_M = 1e5 + 5;
int fac[MAX_N];
int C(int n, int m) {
if (m > n) return 0;
return 1ll * fac[n] * fpow(fac[n - m], Mod - 2) % Mod * fpow(fac[m], Mod - 2) % Mod;
}
int N, M, S, Limit, mx;
int W[MAX_N], A[MAX_N << 2], B[MAX_N << 2], rev[MAX_N << 2];
void NTT(int *p, int op) {
for (int i = 0; i < Limit; i++) if (rev[i] > i) swap(p[rev[i]], p[i]);
for (int i = 1; i < Limit; i <<= 1) {
int rot = fpow(op == 1 ? G : iG, (Mod - 1) / (i << 1));
for (int j = 0, pls = (i << 1); j < Limit; j += pls) {
int w = 1;
for (int k = 0; k < i; k++, w = 1ll * w * rot % Mod) {
int x = p[j + k], y = 1ll * w * p[i + j + k] % Mod;
p[j + k] = (x + y) % Mod, p[i + j + k] = (x - y + Mod) % Mod;
}
}
}
}
int main () {
N = gi(), M = gi(), S = gi();
for (int i = 0; i <= M; i++) W[i] = gi();
mx = max(N, M);
fac[0] = 1; for (int i = 1; i <= mx; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
int mn = min(M, N / S);
Limit = 1; int P = 0;
while (Limit < (mn + 1) << 1) Limit <<= 1, ++P;
for (int i = 0; i <= mn; i++)
A[i] = 1ll * fac[i] * C(M, i) % Mod *
fac[N] % Mod * fpow(M - i, N - i * S) % Mod *
fpow(1ll * fpow(fac[S], i) * fac[N - i * S] % Mod, Mod - 2) % Mod;
for (int i = 0; i <= mn; i++) {
B[i] = fpow(fac[mn - i], Mod - 2);
if ((mn - i) & 1) B[i] = Mod - B[i];
}
for (int i = 0; i < Limit; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (P - 1));
NTT(A, 1), NTT(B, 1);
for (int i = 0; i < Limit; i++) A[i] = 1ll * A[i] * B[i] % Mod;
NTT(A, 0);
int inv = fpow(Limit, Mod - 2);
for (int i = 0; i < Limit; i++) A[i] = 1ll * inv * A[i] % Mod;
int ans = 0;
for (int i = 0; i <= mn; i++) ans = (ans + 1ll * W[i] * A[mn + i] % Mod * fpow(fac[i], Mod - 2) % Mod) % Mod;
printf("%d
", ans);
return 0;
}