HDU 5641:King's Phone【模拟】

King's Phone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 74    Accepted Submission(s): 23

Problem Description

In a military parade, the King sees lots of new things, including an Andriod Phone. He becomes interested in the pattern lock screen.

The pattern interface is a  3×3  square lattice, the three points in the first line are labeled as  1,2,3 , the three points in the second line are labeled as  4,5,6 , and the three points in the last line are labeled as  7,8,9 。The password itself is a sequence, representing the points in chronological sequence, but you should follow the following rules:

- The password contains at least four points.


- Once a point has been passed through. It can't be passed through again.

- The middle point on the path can't be skipped, unless it has been passed through( 3427  is valid, but  3724  is invalid).

His password has a length for a positive integer  k(1≤k≤9) , the password sequence is  s1,s2...sk(0≤si<INT_MAX)  , he wants to know whether the password is valid. Then the King throws the problem to you.

 

Input

The first line contains a number&nbsp; T(0<T≤100000) , the number of the testcases.

For each test case, there are only one line. the first first number&nbsp; k ，represent the length of the password, then  k  numbers, separated by a space, representing the password sequence  s1,s2...sk .

 

Output

Output exactly  T  lines. For each test case, print `valid` if the password is valid, otherwise print `invalid`

 

Sample Input

   
   
   
   

    
    
    
    3
4 1 3 6 2
4 6 2 1 3
4 8 1 6 7
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    invalid
valid
valid

hint:
For test case #1：The path $1\rightarrow 3$ skipped the middle point $2$, so it's invalid.

For test case #2：The path $1\rightarrow 3$ doesn't skipped the middle point $2$, because the point 2 has been through, so it's valid.

For test case #2：The path $8\rightarrow 1 \rightarrow 6 \rightarrow 7$ doesn't have any the middle point $2$, so it's valid.
   
   
   
   

 

考虑要全面。。。。

AC-code：

#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
	int T,n,i,k,a,flag,vis[10],m[10];
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		flag=0;
		for(i=0;i<n;i++)
		{
			scanf("%d",&m[i]);
			if(m[i]<1||m[i]>9)
				flag=1;
		}
		if(n<4||flag)
		{
			printf("invalid\n");
			continue;
		}
		memset(vis,0,sizeof(vis));
		vis[m[0]]=1;
		for(i=0;i<n-1;i++)
		{
			k=m[i];
			a=m[i+1];
			if(vis[a])
				break;
			 vis[k]=vis[a]=1;
			if(k-a==2&&(k==3||k==6||k==9))
			{
				if(!vis[k-1])
					break;
			}
			else if(a-k==2&&(a==3||a==6||a==9))
			{
				if(!vis[a-1])
					break;
			}
			else if(a-k==6&&(a==7||a==8||a==9))
			{
				if(!vis[a-3])
					break;
			}
			else if(k-a==6&&(k==7||k==8||k==9))
			{
				if(!vis[k-3])
					break;
			}
			else if(a+k==10)
				if(!vis[5])
					break;
			else if((a==0&&k==5)||(a==5&&k==0))
				if(!vis[8])
					break;
			else if((a==0&&k==2)||(a==2&&k==0))
				if(!vis[8]||!vis[5])
					break;
		}
		if(i==n-1)
			printf("valid\n");
		else
			printf("invalid\n");
	}
	return 0;
}