HDU 5641 King's Phone 模拟
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;
}
HDU 5641 King's Phone 模拟的更多相关文章
- hdu 5641 King's Phone(暴力模拟题)
Problem Description In a military parade, the King sees lots of new things, including an Andriod Pho ...
- HDU 5641 King's Phone【模拟】
题意: 给定一串密码, 判断是否合法. 长度不小于4 不能重复经过任何点 不能跳过中间点,除非中间点已经经过一次. 分析: 3*3直接记录出可能出现在两点之间的点,直接模拟就好. 注意审题,别漏了判断 ...
- hdu 5641 King's Phone
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5641 题目类型:水题 题目思路:将点x到点y所需要跨过的点存入mark[x][y]中(无需跨过其它点存 ...
- hdu 5641 BestCoder Round #75
King's Phone Accepts: 310 Submissions: 2980 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: ...
- HDU 4041 Eliminate Witches! (模拟题 ACM ICPC 2011亚洲北京赛区网络赛)
HDU 4041 Eliminate Witches! (模拟题 ACM ICPC 2011 亚洲北京赛区网络赛题目) Eliminate Witches! Time Limit: 2000/1000 ...
- hdu 5640 King's Cake(模拟)
Problem Description It is the king's birthday before the military parade . The ministers prepared ...
- HDU 5640 King's Cake【模拟】
题意: 给定长方形,每次从中切去一个最大的正方形,问最终可以得到多少正方形. 分析: 过程类似求gcd,每次减去最小的边即可. 代码: #include <cstdio> #include ...
- POJ 3344 & HDU 2414 Chessboard Dance(模拟)
题目链接: PKU:http://poj.org/problem? id=3344 HDU:http://acm.hdu.edu.cn/showproblem.php?pid=2414 Descrip ...
- HDU 5948 Thickest Burger 【模拟】 (2016ACM/ICPC亚洲区沈阳站)
Thickest Burger Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)T ...
随机推荐
- weblogic nmap扫描脚本
CVE-2018-2894 / Nmap利用脚本,可批量批量快速扫描getshell.检测漏洞.利用漏洞 地址:https://github.com/Rvn0xsy/nse_vuln/tree/ma ...
- C语言回调函数总结
/* Main program ---calls--> Library function ---calls--> Callback funtion */ #include <stdi ...
- SQLite3使用详解
sqlite常量的定义(SQLite3返回值的意思): SQLITE_OK = 0; 返回成功 SQLITE_ERROR = 1; SQL错误或错误的数据库 SQ ...
- python的sorted函数对字典按value进行排序
场景:词频统计时候,我们往往要对频率进行排序 sorted(iterable,key,reverse),sorted一共有iterable,key,reverse这三个参数.其中iterable表示可 ...
- fastJson去掉指定字段
public static String filterFieldsJson(Object src, Class<?> clazz, String... args) { SimpleProp ...
- Windows: 在系统启动时运行程序、定时计划任务、定时关机
lesca今天介绍一些让系统在启动时,而非登录时,加载用户自定义的应用程序或脚本的方法,推荐度从前到后依次递减. 1. Windows任务计划(task scheduler) 用户可以按以下步骤进行操 ...
- Geoserver WFS跨域设置
测试版本为geoserver2.11.0. 两种方法都可以实现跨域设置: 第一种: 下载跨域jar包jetty-servlets.jar(下载geoserver使用的对应jetty版本——可以查看&l ...
- java版云笔记(六)之AOP
今天主要是利用aop技术追加service的响应时间的计算和异常的日志记录. AOP AOP(Aspect Oriented Programming),即面向切面编程,可以说是OOP(Object O ...
- 版本控制软件——tortoiseSVN的基础使用
零 基本功能介绍... 2 一 安装及下载client端... 2 二 登陆和文件下载... 2 三 新增档案及目录到服务器中... 4 四 文件对比... 13 4.1 文件回溯... 13 4.2 ...
- leetcode 之Search in Rotated Sorted Array(四)
描述 Follow up for ”Search in Rotated Sorted Array”: What if duplicates are allowed? Would this aff ...