Codeforces 960 G Bandit Blues

传送门.

题目大意：

求有多少个n的排列：
满足：
1.有a个数比它左边的都大
2.有b个数比它右边的都大

1 <= n <= 10^5

题解：

设$f_{i,j}$表示i的排列，有j个数比它左边的都大的方案数。

考虑加一个最小的数进来，转移为$f_{i,j}=f_{i-1,j-1}+(i-1)*f_{i -1,j}$

发现这就是无符号第一类斯特兰数。

怎么同时满足条件a,b呢？

枚举最大的数的位置，显然a个数都在这个数的左边，b个数都在这个数的右边。

$Ans=\sum_{i=1}^n s_{n-1,a-1}*s_{n-1-i,b-1}*C_{n-1}^{i-1}$

这个瞎合并成：
$Ans=s_{n-1}^{a+b-2}*C_{a+b-2}^{a-1}$

分治NTT去求$s_{n-1}^{a+b-2}$。

Code：

#include<cstdio>
#include<algorithm>
#define ll long long
#define fo(i, x, y) for(int i = x; i <= y; i ++)
#define fd(i, x, y) for(int i = x; i >= y; i --)
#define ff(i, x, y) for(int i = x; i < y; i ++)
using namespace std;

const int mo = 998244353;

int n, A, B;

ll ksm(ll x, ll y) {
    ll s = 1;
    for(; y; y /= 2, x = x * x % mo)
        if(y & 1) s = s * x % mo;
    return s;
}

ll fac(int n) {
    ll s = 1; fo(i, 1, n) s = s * i % mo;
    return s;
}

ll C(int n, int m) {
    return fac(n) * ksm(fac(m) * fac(n - m) % mo, mo - 2) % mo;
}

const int N = 8e5 + 5;

ll d[N]; int p[N], q[N];

ll w[N], a[N], b[N]; int tp;

void dft(ll *a, int n) {
    ff(i, 0, n) {
        int p = i, q = 0;
        fo(j, 1, tp) q = q * 2 + p % 2, p /= 2;
        if(i > q) swap(a[i], a[q]);
    }
    for(int m = 2; m <= n; m *= 2) {
        int h = m / 2;
        ff(i, 0, h) {
            ll W = w[i * (n / m)];
            for(int j = i; j < n; j += m) {
                int k = j + h;
                ll u = a[j], v = a[k] * W % mo;
                a[j] = (u + v) % mo; a[k] = (u - v + mo) % mo;
            }
        }
    }
}

void fft(ll *a, ll *b, int n) {
    ll v = ksm(3, (mo - 1) / n);
    w[0] = 1; ff(i, 1, n) w[i] = w[i - 1] * v % mo ;
    dft(a, n); dft(b, n);
    ff(i, 0, n) a[i] = a[i] * b[i] % mo;
    ff(i, 1, n / 2) swap(w[i], w[n - i]);
    dft(a, n); v = ksm(n, mo - 2);
    ff(i, 0, n) a[i] = a[i] * v % mo;
}

void dg(int x, int y) {
    if(x == y) return;
    int m = x + y >> 1;
    dg(x, m); dg(m + 1, y);
    int n0 = q[x] - p[x] + q[m + 1] - p[m + 1];
    tp = 0; while(1 << ++ tp <= n0); n0 = 1 << tp;
    ff(i, 0, n0) a[i] = b[i] = 0;
    fo(i, p[x], q[x]) a[i - p[x]] = d[i];
    fo(i, p[m + 1], q[m + 1]) b[i - p[m + 1]] = d[i];
    fft(a, b, n0);
    q[x] += q[m + 1] - p[m + 1];
    fo(i, p[x], q[x]) d[i] = a[i - p[x]];
}

ll solve(int n, int m) {
    if(n < m) return 0;
    if(n == 0 && m == 0) return 1;
    fo(i, 0, n - 1) {
        p[i] = ++ tp; q[i] = ++ tp;
        d[p[i]] = i; d[q[i]] = 1;
    }
    dg(0, n - 1);
    return d[p[0] + m];
}

int main() {
    scanf("%d %d %d", &n, &A, &B);
    if(A == 0 || B == 0) {
        printf("0\n"); return 0;
    }
    printf("%I64d", solve(n - 1, A + B - 2) * C(A + B - 2, A - 1) % mo);
}