King’s Phone

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5641

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 T(0<T≤100000), the number of the testcases.

For each test case, there are only one line. the first first number 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.

Hint

题意

手机锁屏

3*3的格子，需要满足下列四个条件：

1.至少4位密码

2.数字没有重复出现

3.经过的位置之间的数字不能跳过，除非之前经过过。

给你一个串序列，问你是否合法。

题解：

模拟题。

有坑，注意每个数是[0,inf)的……

至少四位数。

注意这些，然后瞎写写就好了。

代码

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<math.h>
#include<cstring>
using namespace std;
int a[30];
int vis[30];
int check(int x , int y){
    if( x > y ) swap( x , y );
    if( x == 1 && y == 3) return 2;
    else if( x == 1 && y == 7) return 4;
    else if( x == 1 && y == 9) return 5;
    else if( x == 2 && y == 8) return 5;
    else if( x == 3 && y == 9) return 6;
    else if( x == 3 && y == 7) return 5;
    else if( x == 4 && y == 6) return 5;
    else if( x == 7 && y == 9) return 8;
    return -1;
}

void solve()
{
    int flag = 0;
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    if(n<4||n>9)
    {
        printf("invalid\n");
        return;
    }
    for(int i=1;i<=n;i++)
        if(a[i]<=0||a[i]>9)
        {
            printf("invalid\n");
            return;
        }
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)
    {
        if(vis[a[i]])
        {
            printf("invalid\n");
            return;
        }
        vis[a[i]]++;
    }
    memset(vis,0,sizeof(vis));
    for(int i=1;i<n;i++)
    {
        vis[a[i]]=1;
        int p = check(a[i],a[i+1]);
        if(p!=-1&&vis[p]==0)
        {
            printf("invalid\n");
            return;
        }
    }
    printf("valid\n");
    return;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)solve();
    return 0;
}