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 $3 \times 3$ 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 $k (1\le k\le 9)$, the password sequence is $s_1,s_2...s_k(0\le s_{i} < INT\_MAX)$ , he wants to know whether the password is valid. Then the King throws the problem to you.
Input
The first line contains a number $T(0 < T \le 100000)$, 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 $s_1,s_2...s_k$.
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 $1\rightarrow 3$ 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.
Source
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
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 #include<map>
5 #include<queue>
6 #include<stack>
7 #include<set>
8 using namespace std;
9 int t;
10 int a[10];
11 int mp[10][10];
12 map<int,int> mpp;
13 int jishu=0;
14 void init()
15 {
16 mp[1][3]=1;
17 mp[1][7]=1;
18 mp[1][9]=1;
19 mp[2][8]=1;
20 mp[3][1]=1;
21 mp[3][9]=1;
22 mp[3][7]=1;
23 mp[4][6]=1;
24 mp[6][4]=1;
25 mp[7][1]=1;
26 mp[7][9]=1;
27 mp[7][3]=1;
28 mp[8][2]=1;
29 mp[9][3]=1;
30 mp[9][7]=1;
31 mp[9][1]=1;
32 }
33 int main()
34 {
35 scanf("%d",&t);
36
37 init();
38 for(int i=1;i<=t;i++)
39 {
40 scanf("%d",&jishu);
41 memset(a,0,sizeof(a));
42 int maxn=0;
43 int flag=1;
44 for(int j=1;j<=jishu;j++)
45 {
46 scanf("%d",&a[j]);
47 if(a[j]>maxn)
48 maxn=a[j];
49 if(a[j]<=0)
50 flag=0;
51 }
52 mpp.clear();
53 if(jishu<4||jishu>9||maxn>9||flag==0)
54 printf("invalid\n");
55 else
56 { int re=0;
57 mpp[a[1]]=1;
58 for(int j=1;j<jishu;j++)
59 {
60 if(mpp[a[j+1]]==0)
61 {
62 if(mp[a[j]][a[j+1]]==0)
63 re++;
64 else
65 {
66 if(mpp[(a[j]+a[j+1])/2])
67 re++;
68 }
69 }
70 mpp[a[j+1]]=1;
71
72 }
73 if(re==jishu-1)
74 printf("valid\n");
75 else
76 printf("invalid\n");
77 }
78 }
79 return 0;
80 }