CF960G Bandit Blues
第300篇blog yeah
https://www.luogu.com.cn/problem/CF960G
这题是P4609 [FJOI2016]建筑师的加强版
一眼秒了，没有什么好说的
考虑
n
n
n会把序列分成左右两个部分
对于每个建筑，把它和被它挡住的放在一个集合，假设这个集合大小为sz，则方案数为
(
s
z
−
1
)
!
(sz-1)!
(sz−1)!除了最高那个在最前面，后面的可以随便交换位置，容易发现这就是圆排列，进而得出第一类斯特林数
那么答案显然就是
[
n
−
1
A
−
1
+
B
−
1
]
×
(
A
−
1
+
B
−
1
A
−
1
)
\begin{bmatrix}n-1 \\ A-1+B-1\end{bmatrix} \times \binom{A-1+B-1}{A-1}
[n−1A−1+B−1]×(A−1A−1+B−1)
第一类斯特林数求一项并不好求，所以可以直接大力把一行求出来，取对应项即可
code:
#include<bits/stdc++.h>
#define mod 998244353
#define N 4000500
using namespace std;
int add(int x, int y) { x += y;
if(x >= mod) x -= mod;
return x;
}
int sub(int x, int y) { x -= y;
if(x < 0) x += mod;
return x;
}
int mul(int x, int y) {
return 1ll * x * y % mod;
}
int qpow(int x, int y) {
int ret = 1;
for(; y; y >>= 1, x = mul(x, x)) if(y & 1) ret = mul(ret, x);
return ret;
}
const int G = 3;
const int Ginv = qpow(G, mod - 2);
int rev[N];
void ntt(int *a, int n, int o) {
for(int i = 1; i < n; i ++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * (n >> 1));
for(int i = 1; i < n; i ++) if(i > rev[i]) swap(a[i], a[rev[i]]);
for(int len = 2; len <= n; len <<= 1) {
int w0 = qpow(o == 1? G : Ginv, (mod - 1) / len);
for(int j = 0; j < n; j += len) {
int wn = 1;
for(int k = j; k < j + (len >> 1); k ++, wn = mul(wn, w0)) {
int X = a[k], Y = mul(a[k + (len >> 1)], wn);
a[k] = add(X, Y); a[k + (len >> 1)] = sub(X, Y);
}
}
}
int ninv = qpow(n, mod - 2);
if(o == -1)
for(int i = 0; i < n; i ++) a[i] = mul(a[i], ninv);
}
int n, fac[N], ifac[N], a[N], b[N], f[N], g[N];
void init(int n) {
fac[0] = 1;
for(int i = 1; i <= n; i ++) fac[i] = mul(fac[i - 1], i);
ifac[n] = qpow(fac[n], mod - 2);
for(int i = n - 1; i >= 0; i --) ifac[i] = mul(ifac[i + 1], i + 1);
}
void Mul(int *a, int *b, int n, int m) {
int len = 1;
for(; len <= n + m; ) len <<= 1;
ntt(a, len, 1), ntt(b, len, 1);
for(int i = 0; i < len; i ++) a[i] = mul(a[i], b[i]);
ntt(a, len, - 1);
for(int i = n + m + 1; i <= len; i ++) a[i] = 0;
}
void solve(int *f, int n) {
if(n == 0) {f[0] = 1; return ;}
if(n == 1) {f[1] = 1; return ;}
if(n & 1) {
solve(f, n - 1);
for(int i = n; i >= 1; i --) {
f[i] = add(f[i - 1], mul(f[i], n - 1));
}
f[0] = mul(f[0], n - 1);
} else {
solve(f, n >> 1);
int m = n / 2, mi = 1;
for(int i = 0; i <= m; i ++, mi = mul(mi, m))
a[i] = mul(f[i], fac[i]), b[i] = mul(mi, ifac[i]);
reverse(a, a + 1 + m);
Mul(a, b, m, m);
for(int i = 0; i <= m; i ++) g[i] = mul(a[m - i], ifac[i]);
Mul(f, g, m, m);
int len = 1;
for( ;len <= m + m; ) len <<= 1;
for(int i = 0; i <= len; i ++) g[i] = a[i] = b[i] = 0;
// for(int i = 0; i <= n; i ++) printf("%d ", f[i]); printf(" %d\n", n);
}
}
int C(int n, int m) {
return 1ll * fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
int A, B;
int main() {
scanf("%d%d%d", &n, &A, &B);
init(n);
n --;
solve(f, n);
printf("%lld", 1ll * f[A + B - 2] * C(A + B - 2, A - 1) % mod);
return 0;
}
