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.

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)

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

2
1 1
3
1 1 1
2
1000000 1

1
0
0

2012 ACM/ICPC Asia Regional Changchun Online

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<stack>
#include<vector>
#include<map>

using namespace std;

int n;
__int64 a[10010];
int v[10010];
__int64 b[10010];

int main() {
    while(scanf("%d",&n)!=EOF) {
        for(int i=1; i<=n; i++) {
            scanf("%I64d",&a[i]);
        }
        memset(v,0,sizeof(v));
        int t = 0;
        int flag;
        for(int i=1; i<=n; i++) {
            int pt = t;
            flag = 0;
            if(v[i] == 1){
				continue;
            }
            for(int k=1; k<=t; k++) {
                if(a[i] == b[k]) {
                    for(int pi=k; pi<t; pi++) {
                        b[pi] = b[pi+1];
                    }
                    t--;
                    flag = 1;
                }
            }
            if(pt == t) {
                int pf = 0;
                for(int j=i+1; j<=i+5 && j<=n; j++) {
                    if(a[i] == a[j]) {
                        v[i] = 1;
                        v[j] = 1;
                        pf = 1;
                        break;
                    }
                }
                if(pf == 0) {
                    b[++t] = a[i];
                }
            }
        }
        if(t == 0) {
            printf("1\n");
        } else {
            printf("0\n");
        }
    }
    return 0;
}