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 (GCC 5.4.1) (judge verdict: Accepted)


#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>

int main(void)
{
	//変数の宣言
	int n;
	int d[60];
	int i,j,k;
	int time[13];
	int flag=0;
	int time24[25];
	int temp=0;
	int ans=100;
	int lastans=0;
	
	//データの読み込み
	scanf("%d",&n);
	
	
	for(i=0;i<=12;i++){
		time[i]=0;
	}
	
	for(i=0;i<24;i++){
		time24[i]=0;
	}
	for(i=0;i<n;i++){
		scanf("%d",&d[i]);
		time[d[i]]++;
	}
	
//	printf("nは%dです\n", n);
//	printf("データの読み込み終了\n");
	//実際の処理
	
	if(n>=24){
		printf("0");
		return 0;
	}else if(n<12){
//		printf("n<12\n");
		for(i=0;i<=12;i++){
			if(i==0){
				if(time[i]>=1) flag=1;
			}else if(i==12){
				if(time[i]>=2) flag=1;
			}else{
				if(time[i]>=3) flag=1;
			}
		}
		if(flag==1){
			printf("0");
			return 0;
		}
		
//		printf("s!=0\n");
		
		time24[0]=1;
		time24[24]=1;
		for(i=0;i<(1<<n);i++){
//			printf("i=%d\n",i);
			
			for(j=1;j<24;j++){
				time24[j]=0;
			}
			
			for(j=0;j<n;j++){
				if(i&(1<<j)){
					if(time24[d[j]]==0){
						time24[d[j]]=1;
					}else{
						time24[24-d[j]]=1;
					}
				}else{
					if(time24[24-d[j]]==0){
						time24[24-d[j]]=1;
					}else{
						time24[d[j]]=1;
					}
				}
			}
			
//			for(j=0;j<24;j++){
//				printf("time24[%d]=%d\n",j,time24[j]);
//			}
			temp=1;
			ans=100;
			for(j=1;j<=24;j++){
				if(time24[j]==1){
					if(temp<ans) ans=temp;
//					printf("temp=%d ans=%d",temp,ans);
					temp=1;
				}else{
					temp++;
				}
			}
			if(lastans<ans) lastans=ans;
//			printf("%d\n",ans);
		}
		printf("%d",lastans);
	}else{
		for(i=0;i<=12;i++){
			if(i==0){
				if(time[i]>=1) flag=1;
			}else if(i==12){
				if(time[i]>=2) flag=1;
			}else{
				if(time[i]>=3) flag=1;
			}
		}
		if(flag==1){
			printf("0");
		}else{
			printf("1");
		}
		return 0;
	}
	
//	printf("計算部分終了\n");
	
	//出力
	
//	printf("結果の出力終了\n");
	
}