2012 ACM/ICPC Asia Regional Changchun Online-LianLianKan

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

21 131 1 121000000 1

 

Sample Output

100
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;}