LianLianKan

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2389    Accepted Submission(s): 382

Problem Description

I like playing game with my friends, although sometimes look 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 the 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 wisdom among my friends, 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 5.

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 integers 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

题意：类似于我们玩的连连看，从上往下，6个水果内如果有相同的2个水果则可以消去，直至水果被消完，输出1，或是找不到可以消去的水果，输出0。

思路：数据顶多1000个，直接暴力过。

#include<stdio.h>
int main()
{
    int a[1111];
    int n,i,j;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        if(n%2) //因为每次只能消去2个，所以奇数个水果必然消不完
        {
            printf("0\n");
            continue;
        }
        int flag=n;
        while(flag>0)
        {
            int hash=0;
            for(i=0;i<n;i++)
            {
                if(a[i]!=-1)
                {
                    int num=0;
                    for(j=i+1;j<n;j++)
                    {
                        if(num>4) //6个水果内找不到2个相同的
                            break;
                        if(a[j]==a[i])
                        {
                            a[j]=-1; //消去的水果进行标记
                            a[i]=-1; //消去的水果进行标记
                            hash=1; //输出标记
                            flag-=2; //总水果数对应的减去2
                            break;
                        }
                        else
                        {
                            if(a[j]!=-1)
                                num++;
                        }
                    }
                }
                if(hash==1)
                    break;
            }
            if(hash==0)
            {
                printf("0\n");
                break;
            }
        }
        if(flag==0)
            printf("1\n");
    }
    return 0;
}