https://blog.csdn.net/zxyoi_dreamer/article/details/105273205
https://www.cnblogs.com/Master-Yoda/p/12547799.html
我还是没太搞懂zxyoi dalao中(g)数组怎么求的，还是最好用Master-Yoda的方法吧
#include<bits/stdc++.h>
using namespace std;
#define fp(i,l,r) for(register int (i)=(l);(i)<=(r);++(i))
#define fd(i,l,r) for(register int (i)=(l);(i)>=(r);--(i))
#define fe(i,u) for(register int (i)=front[(u)];(i);(i)=e[(i)].next)
#define mem(a) memset((a),0,sizeof (a))
#define O(x) cerr<<#x<<':'<<x<<endl
#define int long long
inline int read(){
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return x*f;
}
void wr(int x){
if(x<0)putchar('-'),x=-x;
if(x>=10)wr(x/10);
putchar('0'+x%10);
}
const int MAXN=601,mod=1e9+7,inv2=(mod+1)/2;
inline void tmod(int &x){x%=mod;}
int n,dp[MAXN],g[MAXN],C[MAXN][MAXN],s0,s1;
bool vis[MAXN*2];
main(){
n=read();
fp(i,1,n)vis[read()]=1;
fp(i,0,n){
C[i][0]=1;
fp(j,1,i)tmod(C[i][j]=C[i-1][j]+C[i-1][j-1]);
}
g[0]=1;
fp(i,1,n)fp(j,0,i-1)
tmod(g[i]+=g[j]*g[i-j-1]%mod*C[i-1][j]%mod*(j+2));
dp[0]=1;
fd(i,n*2,1){
if(vis[i]){
fd(j,s1,0)if(dp[j])fp(k,0,s1-j)
tmod(dp[j+k+1]+=dp[j]*g[k]%mod*C[s1-j][k]%mod*(k+2));
++s1;
}
else{
fp(j,0,s1)tmod(dp[j]*=j-s0);
++s0;
}
}
int ans=dp[n];
fp(i,1,n)tmod(ans*=inv2);
printf("%lld
",ans);
}