Problem:
Score : 500 points

Problem Statement

In CODE FESTIVAL XXXX, there are N+1 participants from all over the world, including Takahashi.

Takahashi checked and found that the time gap (defined below) between the local times in his city and the i-th person's city was D_i hours.
The time gap between two cities is defined as follows. For two cities A and B, if the local time in city B is d o'clock at the moment when the local time in city A is 0 o'clock, then the time gap between these two cities is defined to be min(d,24-d) hours.
Here, we are using 24-hour notation.
That is, the local time in the i-th person's city is either d o'clock or 24-d o'clock at the moment when the local time in Takahashi's city is 0 o'clock, for example.

Then, for each pair of two people chosen from the N+1 people, he wrote out the time gap between their cities. Let the smallest time gap among them be s hours.

Find the maximum possible value of s.

Constraints

1 \leq N \leq 50

0 \leq D_i \leq 12

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
D_1 D_2 ... D_N

Output

Print the maximum possible value of s.

Sample Input 1

3
7 12 8

Sample Output 1

4

For example, consider the situation where it is 7, 12 and 16 o'clock in each person's city at the moment when it is 0 o'clock in Takahashi's city. In this case, the time gap between the second and third persons' cities is 4 hours.

Sample Input 2

2
11 11

Sample Output 2

2

Sample Input 3

1
0

Sample Output 3

0

Note that Takahashi himself is also a participant.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<memory.h>

using namespace std;
#define REP(i,m,n) for(int i=(int)m ; i < (int) n ; ++i )
#define rep(i,n) REP(i,0,n)
typedef long long ll;
typedef pair<int,int> P;
typedef pair<ll,int> pli;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=1e9+7 ;

int cnt[100];

int main(){
  int n;
  cin >> n;
  int d[n];
  rep(i,n)cin >> d[i];
  if(n>=25){
    cout << 0 << endl;
    return 0;
  }
  if(n==24){
    int cnt[30]={};
    rep(i,n){
      cnt[d[i]-1]++;
    }
    rep(i,12){
      if(cnt[i]<2){
        cout << 0 << endl;
        return 0;
      }
    }
    cout << 1 << endl;
    return 0;
  }
  int ans=-1;
  for(int bit=0;bit<(1<<n);bit++){
    int temp[n]={};
    int res=inf;
    for(int i=0;i<n;i++){
      if(bit & (1<<i) && d[i]!=0){
        temp[i]=24-d[i];
      }else{
        temp[i]=d[i];
      }
    }
    sort(temp,temp+n);
    rep(i,n-1){
      res=min(res,temp[i+1]-temp[i]);
    }
    res=min(res,temp[0]);
    ans=max(ans,res);
  }
  cout << ans << endl;
return 0;}