Problem Description

I like playing game with my friend, although sometimes looks pretty naive. Today I invent a new game called LianLianKan. The game is about playing on a number stack.
Now we have a number stack, and we should link and pop the same element pairs from top to bottom. Each time, you can just link the top element with one same-value element. After pop them from stack, all left elements will fall down. Although the game seems to be interesting, it's really naive indeed. 

To prove I am a wisdom among my friend, I add an additional rule to the game: for each top element, it can just link with the same-value element whose distance is less than 6 with it. 
Before the game, I want to check whether I have a solution to pop all elements in the stack.

 

Input

There are multiple test cases.
The first line is an integer N indicating the number of elements in the stack initially. (1 <= N <= 1000)
The next line contains N integer ai indicating the elements from bottom to top. (0 <= ai <= 2,000,000,000)

 

Output

For each test case, output “1” if I can pop all elements; otherwise output “0”.

 

Sample Input

 

Sample Output

 

Source

 #include<stdio.h>

 #include<iostream>

 #include<string>

 #include<cstring>

 #include<algorithm>

 #include<queue>

 #include<set>

 #include<map>

 using namespace std;

 int vis[],a[];

 int dfs(int n)

 {

     int i=,j=n-;

     while(n>=&&vis[n]) --n;

     if(n==-) return ;

     if(n==&&!vis[]) return ;

     while(i<=)

     {

         if(j<) return ;

         if(vis[j]) --j;

         if(a[n]==a[j])

         {

             vis[j]=;

             if(dfs(n-)) return ;

             vis[j]=;

         }

         ++i;

         --j;

     }

     return ;

 }

 int main()

 {

     int k,m,q,t,p,n;

     int T;

     map<int,int> mp;

     map<int,int>::iterator it;

     while(cin>>n)

     {

         t=;

         for(int i=;i<n;++i)

         {

             scanf("%d",&a[i]);

             vis[i]=;

             mp[a[i]]++;

         }

         for(it=mp.begin();it!=mp.end();++it)

         {

             if((*it).second%)

             {

                 t=;

                 break;

             }

         }

         if(t)

         {

             cout<<<<endl;

         }else

         {

             cout<<dfs(n-)<<endl;

         }

     }

     return ;

 }