PLEASE

思路

• 010会分出两种情况100和001
  100会分出两种情况100和010
  001会分出两种情况100和001
  

• 经过i次交换后，可以得到个010
  
  
  当n为偶数，
  当n为奇数，

代码

// Problem: PLEASE
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/problem/111563
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp(aa,bb) make_pair(aa,bb)
#define _for(i,b) for(int i=(0);i<(b);i++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,b,a) for(int i=(b);i>=(a);i--)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define X first
#define Y second
#define lowbit(a) (a&(-a))
#define debug(a) cout<<#a<<":"<<a<<"\n"
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
typedef long double ld;
const int N=100010;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const double eps=1e-6;
const double PI=acos(-1.0);

ll fpow(ll a,ll b){
    if(mod==1) return 0;
    ll ans=1%mod;
    while(b){
        if(b&1) ans=ans*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ans;
}


void solve(){
    bool ou=0;
    ll t=2; //t=2^(n-1)
    int n;cin>>n;
    rep(i,1,n){
        ll a;cin>>a;
        t=fpow(t,a);
        if(a%2==0) ou=1;
    }
    t=t*fpow(2,mod-2)%mod;
    if(ou) cout<<(t+1)*fpow(3,mod-2)%mod<<"/"<<t<<"\n";
    else cout<<(t-1+mod)%mod*fpow(3,mod-2)%mod<<"/"<<t<<"\n";
}


int main(){
    ios::sync_with_stdio(0);cin.tie(0);
//    int t;cin>>t;while(t--)
    solve();
    return 0;
}