BZOJ5322 JXOI2018排序问题

　　对于一个序列，重排后有序的概率显然是∏cnt_i!/n!，其中cnt_i为第i种数出现次数。要使概率最小，显然应该让各种数字尽量平均分配。剩下的是div2BC左右的大讨论。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 200010
#define M 10000010
#define P 998244353
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int T,n,m,l,r,a[N],cnt[N],fac[N+M];
int ksm(int a,int k)
{
    int s=;
    for (;k;k>>=,a=1ll*a*a%P) if (k&) s=1ll*s*a%P;
    return s;
}
int inv(int a){return ksm(a,P-);}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj5322.in","r",stdin);
    freopen("bzoj5322.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    T=read();
    fac[]=;for (int i=;i<N+M;i++) fac[i]=1ll*fac[i-]*i%P;
    while (T--)
    {
        n=read(),m=read(),l=read(),r=read();int u=n,v=m;
        for (int i=;i<=n;i++) a[i]=read();
        sort(a+,a+n+);
        int tot=;
        for (int i=;i<=n;i++)
        {
            int t=i;
            while (t<n&&a[t+]==a[i]) t++;
            cnt[++tot]=t-i+;i=t;
        }
        unique(a+,a+n+);n=tot;
        int L=n+,R=;
        for (int i=n;i>=;i--) if (l<=a[i]) L=i;else break;
        for (int i=;i<=n;i++) if (r>=a[i]) R=i;else break;
        int ans=;
        for (int i=;i<=L-;i++) ans=1ll*ans*fac[cnt[i]]%P;
        for (int i=R+;i<=n;i++) ans=1ll*ans*fac[cnt[i]]%P;
        for (int i=L;i<=R;i++) cnt[i-L+]=cnt[i];
        n=R-L+;sort(cnt+,cnt+n+);tot=r-l+-n;cnt[++n]=P;
        for (int i=;i<=n;i++)
        if (m>=1ll*tot*(cnt[i]-cnt[i-])) m-=tot*(cnt[i]-cnt[i-]),tot++;
        else
        {
            ans=1ll*ans*ksm(fac[cnt[i-]+m/tot],tot)%P*ksm(cnt[i-]+m/tot+,m%tot)%P,m=;
            for (int j=i;j<n;j++) ans=1ll*ans*fac[cnt[j]]%P;
            break;
        }
        ans=1ll*fac[u+v]*inv(ans)%P;
        printf("%d\n",ans);
    }
    return ;
}