HDU 5641

King's Phone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 916    Accepted Submission(s): 261

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 $3 imes 3$ 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 $k (1le kle 9)$, the password sequence is $s_1,s_2...s_k(0le s_{i} < INT\_MAX)$ , 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;$T(0 < T le 100000)$, 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 $s_1,s_2...s_k$.

 

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 $1 ightarrow 3$ skipped the middle point $2$, so it's invalid. For test case #2：The path $1 ightarrow 3$ doesn't skipped the middle point $2$, because the point 2 has been through, so it's valid. For test case #2：The path $8 ightarrow 1 ightarrow 6 ightarrow 7$ doesn't have any the middle point $2$, so it's valid.

 

Source

BestCoder Round #75

挂终测了  继续掉分 一回解  orzzzz 最近 太忙

题意： 给你一段序列 问是否为合法的解锁序列（模拟手机手势解锁）

 

满足条件 1.至少4个数字 数量为[4,9]  2.不能重复 3.不能跳跃 4.数字的范围为[1,9]

还有记得每次的初始化

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<queue>
#include<stack>
#include<set>
using namespace std;
int t;
int a[10];
int mp[10][10];
map<int,int> mpp;
int jishu=0;
void init()
{
    mp[1][3]=1;
    mp[1][7]=1;
    mp[1][9]=1;
    mp[2][8]=1;
    mp[3][1]=1;
    mp[3][9]=1;
    mp[3][7]=1;
    mp[4][6]=1;
    mp[6][4]=1;
    mp[7][1]=1;
    mp[7][9]=1;
    mp[7][3]=1;
    mp[8][2]=1;
    mp[9][3]=1;
    mp[9][7]=1;
    mp[9][1]=1;
}
int main()
{
    scanf("%d",&t);

    init();
    for(int i=1;i<=t;i++)
    {
         scanf("%d",&jishu);
         memset(a,0,sizeof(a));
         int maxn=0;
         int flag=1;
         for(int j=1;j<=jishu;j++)
            {
                scanf("%d",&a[j]);
                if(a[j]>maxn)
                    maxn=a[j];
                if(a[j]<=0)
                    flag=0;
            }
        mpp.clear();
        if(jishu<4||jishu>9||maxn>9||flag==0)
            printf("invalid
");
        else
        {   int re=0;
            mpp[a[1]]=1;
            for(int j=1;j<jishu;j++)
            {
                if(mpp[a[j+1]]==0)
               {
                 if(mp[a[j]][a[j+1]]==0)
                    re++;
                else
                {
                    if(mpp[(a[j]+a[j+1])/2])
                        re++;
                }
               }
                mpp[a[j+1]]=1;

            }
            if(re==jishu-1)
                printf("valid
");
            else
                printf("invalid
");
        }
    }
    return 0;
}