明明的烦恼简化版。
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#define inf 0x7fffffff
#define ll long long
using namespace std;
inline int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int n,m,tot,cnt,d[151],num[151],pri[151];
ll s[25],ans=1;
inline bool jud(int x){
for(int i=2;i<=sqrt(x);++i)
if(x%i==0)return 0;
return 1;
}
inline void getpri(){
for(int i=2;i<=150;++i)
if(jud(i))pri[++cnt]=i;
}
inline void getfac(){
s[1]=1;
for(int i=2;i<=22;++i)
s[i]=s[i-1]*i;
}
inline void solve(ll x,int f){
for(int i=1;i<=cnt;++i){
if(x<=1)return;
while(x%pri[i]==0){
num[i]+=f;
x/=pri[i];
}
}
}
int main(){
getfac();
getpri();
n=read();
if(n==1){
int x=read();
if(!x)printf("1");
else printf("0");
return 0;
}
for(int i=1;i<=n;++i){
d[i]=read();
if(!d[i]){printf("0");return 0;}
tot+=--d[i];
}
if(tot!=n-2){printf("0");return 0;}
solve(s[n-2],1);
for(int i=1;i<=n;++i)
solve(s[d[i]],-1);
for(int i=1;i<=cnt;++i)
while(num[i]--)
ans*=pri[i];
printf("%lld",ans);
return 0;
}