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
2
1 1
3
1 1 1
2
1000000 1


Sample Output
1
0

0

題意分析：本題大意是給你一個棧，棧裡面有有一些數，現從棧頂開始在5個範圍內，找有沒有相同的數，如有相同則消去這兩個數，依次重複 ，如果最後沒有數則輸出1，反之則輸出0.

本題可以用dfs外加map來做，程式碼如下（借鑑了飄哥的程式碼）

#include <iostream>
#include <cstdio>
#include <map>
#include <cstring>

using namespace std;

map<int ,int> m;
int a[1100];
int vis[1100];

int dfs(int n)
{
      while(n && vis[n])
            n--;
      if(n <= 0)
            return 1;
      if(n == 1)
            return 0;
      int i=0, j = n-1;
      for(;i<=5;)
      {
            if(j <= 0)
                  return 0;
            if(vis[j]){
                  j--;
                  continue;
            }
            if(a[n] == a[j])
            {
                  vis[j] = 1;
                  if(dfs(n-1))
                        return 1;
                  vis[j] = 0;
            }
            i++;
            j--;
      }
      return 0;
}

int main()
{
      int n;
      while(scanf("%d",&n)!=EOF)
      {
            m.clear();
            memset(vis,0,sizeof(vis));
            for(int i=1;i<=n;i++)
            {
                  scanf("%d",&a[i]);
                  vis[i] = 0;
                  m[a[i]]++;
            }
            if(n & 1)
            {
                  printf("0\n");
                  continue;
            }
            int flag = 1;
            map<int ,int>::iterator it;
            for(it = m.begin();it != m.end();it++)
            {
                  if(it->second % 2==1)
                  {
                        flag = 0;
                        break;
                  }
            }
            if(flag == 0){
                  printf("0\n");
                  continue;
            }
            printf("%d\n",dfs(n));
      }
      return 0;
}
 

當然本題用另外一種方法也能過，但原理不對（能過是因為杭電資料太水），借鑑了羅明俊俊哥的程式碼：

#include <cstdio>  
using namespace std;  
  
int main(){  
    int n, v, ans, cnt;  
    while(scanf("%d", &n) != EOF){  
        ans = cnt = 0;  
        for(int i = 0; i < n; i++){  
            scanf("%d", &v);  
            ans ^= v;  
            if(v == 0) cnt++;  
        }  
        if(ans == 0)  
            printf("1\n");  
        else  
            printf("0\n");  
    }  
    return 0;  
}  
 

過不了這組資料：

12
1 3 3 5 5 4 4 1 2 6 6 2

但是他所用的方法很值得我們借鑑（異或的自反性），實質上是nim博弈思想的體現。