Divide by Zero 2018 and Codeforces Round #474 (Div. 1 + Div. 2, combined)G - Bandit Blues

题意:求满足条件的排列,1:从左往右会遇到a个比当前数大的数,(每次遇到更大的数会更换当前数)2.从右往左会遇到b个比当前数大的数.
题解:1-n的排列,n肯定是从左往右和从右往左的最后一个数.
考虑\(S(n,m)\)是1-n排列中从左往右会遇到m个比当前数大的数,考虑把1放在最左边,即\(S(n-1,m-1)\),考虑1不在最左边,有n-1个位置,1不可能会更换\((n-1)*S(n,m)\).即\(S(n,m)=S(n-1,m-1)+(n-1)*S(n-1,m)\)
\(S(n,m)\)即第一类斯特林数.答案即\(S(n-1,a+b-2)*C(a+b-2,a-1)\)
\(S(n,*)\)的生成函数即\(\prod_{i=0}^{n-1}(x+i)\),即x的n次上升幂.
\(F_n(x)=\prod_{i=0}^{n-1}(x+i)\),\(F_n(x+n)=\prod_{i=0}^{n-1}(x+n+i)\)
\(F_{2n}(x)=F_n(x)*F_n(x+n)\)
\(F_n(x)=\sum_{i=0}^{n-1}a_ix^i\),\(F_n(x+n)=\sum_{i=0}^{n-1}x^i*\sum_{j=i}^{n-1}\frac{j!}{i!*(j-i)!}n^{j-i}a_j\)
先卷积出\(F_n(x+n)\),然后卷积出\(F_{2n}(x)\),当n不能整除2时,单独考虑乘(x+n-1).
递归处理,复杂度\(O(nlogn)\)

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
//#include <bits/extc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define mt make_tuple
//#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 998244353
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
#define bpc __builtin_popcount
#define base 1000000000000000000ll
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
#define mr mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll mul(ll a,ll b,ll c){return (a*b-(ll)((ld)a*b/c)*c+c)%c;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=mul(ans,a,c);a=mul(a,a,c),b>>=1;}return ans;}

using namespace std;
//using namespace __gnu_pbds;

const ld pi = acos(-1);
const ull ba=233;
const db eps=1e-5;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+10,maxn=2000000+10,inf=0x3f3f3f3f;

ll x[N<<3],y[N<<3];
int rev[N<<3];
void getrev(int bit)
{
    for(int i=0;i<(1<<bit);i++)
        rev[i]=(rev[i>>1]>>1) | ((i&1)<<(bit-1));
}
void ntt(ll *a,int n,int dft)
{
    for(int i=0;i<n;i++)
        if(i<rev[i])
            swap(a[i],a[rev[i]]);
    for(int step=1;step<n;step<<=1)
    {
        ll wn=qp(3,(mod-1)/(step*2));
        if(dft==-1)wn=qp(wn,mod-2);
        for(int j=0;j<n;j+=step<<1)
        {
            ll wnk=1;
            for(int k=j;k<j+step;k++)
            {
                ll x=a[k];
                ll y=wnk*a[k+step]%mod;
                a[k]=(x+y)%mod;a[k+step]=(x-y+mod)%mod;
                wnk=wnk*wn%mod;
            }
        }
    }
    if(dft==-1)
    {
        ll inv=qp(n,mod-2);
        for(int i=0;i<n;i++)a[i]=a[i]*inv%mod;
    }
}
ll f[N],inv[N];
ll C(int a,int b)
{
    if(a<b||a<0||b<0)return 0;
    return f[a]*inv[b]%mod*inv[a-b]%mod;
}
vi solve(int n)
{
    if(n==1)
    {
        vi v={0,1};
        return v;
    }
    vi te=solve(n/2);
    int sz=0,m=te.size();
    while((1<<sz)<m)sz++;sz++;
    int len=1<<sz;
    getrev(sz);
    ll p=1;
    for(int i=0;i<m;i++)
    {
        x[i]=qp(n/2,m-i-1)*inv[m-i-1]%mod;
        y[i]=te[i]*f[i]%mod;
    }
    for(int i=m;i<len;i++)x[i]=y[i]=0;
    ntt(x,len,1);ntt(y,len,1);
    for(int i=0;i<len;i++)x[i]=x[i]*y[i]%mod;
    ntt(x,len,-1);
    for(int i=0;i<m;i++)y[i]=x[i+m-1]*inv[i]%mod;
    for(int i=0;i<m;i++)x[i]=te[i];
    for(int i=m;i<len;i++)x[i]=y[i]=0;
    ntt(x,len,1);ntt(y,len,1);
    for(int i=0;i<len;i++)x[i]=x[i]*y[i]%mod;
    ntt(x,len,-1);
    vi v;v.resize(len+1);
    if(n&1)
    {
        y[0]=n-1;y[1]=1;
        for(int i=0;i<len;i++)
            for(int j=0;j<2;j++)
            {
                v[i+j]+=x[i]*y[j]%mod;
                if(v[i+j]>=mod)v[i+j]-=mod;
            }
    }
    else
    {
        for(int i=0;i<len;i++)v[i]+=x[i];
    }
    while(v.size()&&v.back()==0)v.pop_back();
    return v;
}
int main()
{
    f[0]=inv[0]=1;
    for(int i=1;i<N;i++)f[i]=f[i-1]*i%mod,inv[i]=inv[i-1]*qp(i,mod-2)%mod;
    int n,a,b;scanf("%d%d%d",&n,&a,&b);
    if(a+b-1>n)return 0*puts("0");
    if(n==1)
    {
        if(a==b&&a==1)puts("1");
        else puts("0");
        return 0;
    }
    vi v=solve(n-1);
    printf("%lld\n",1ll*v[a+b-2]*C(a+b-2,a-1)%mod);
    return 0;
}
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原文地址：https://www.cnblogs.com/acjiumeng/p/11567341.html