【暑假】[数学]UVa 1262 Password
UVa 1262 Password
题目:
Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
Suppose that we have a smart phone. If we touch the screen keyboard directly to enter the password, this is very vulnerable since a shoulder-surfer easily knows what we have typed. So it is desirable to conceal the input information to discourage shoulder-surfers around us. Let me explain one way to do this.
You are given a 6 x 5 grid. Each column can be considered the visible part of a wheel. So you can easily rotate each column wheel independently to make password characters visible. In this problem, we assume that each wheel contains the 26 upper letters of English alphabet. See the following Figure 1.
Assume that we have a length-5 password such as p1 p2 p3 p4 p5. In order to pass the authentication procedure, we should construct a configuration of grid space where each pi appears in the i-th column of the grid. In that situation we say that the user password is accepted.
Let me start with one example. Suppose that our password was set `COMPU'. If we construct the grid as shown in Figure 2 on next page, then the authentication is successfully processed.
In this password system, the position of each password character in each column is meaningless. If each of the 5 characters in p1 p2 p3 p4 p5 appears in the corresponding column, that can be considered the correct password. So there are many grid configurations allowing one password. Note that the sequence of letters on each wheel is randomly determined for each trial and for each column. In practice, the user is able to rotate each column and press ``Enter" key, so a should-surfer cannot perceive the password by observing the 6 x 5 grid since there are too many password candidates. In this 6 x 5 grid space, maximally 65 = 7, 776 cases are possible. This is the basic idea of the proposed password system against shoulder-surfers.
Unfortunately there is a problem. If a shoulder-surfer can observe more than two grid plate configurations for a person, then the shoulder-surfer can reduce the searching space and guess the correct password. Even though it is not easy to stealthily observe other's more than once, this is one weakness of implicit grid passwords.
Let me show one example with two observed configurations for a grid password. The user password is `COMPU', but `DPMAG' is also one candidate password derived from the following configuration.
You are given two configurations of grid password from a shoulder-surfer. Suppose that you have succeeded to stealthily record snapshots of the target person's device (e.g. smart phone). Then your next task is to reconstruct all possible passwords from these two snapshots. Since there are lots of password candidates, you are asked for the k-th password among all candidates in lexicographical order. In Figure 3, let us show the first 5 valid password. The first 5 valid passwords are `ABGAG' , `ABGAS', `ABGAU', `ABGPG' and `ABGPS'.
The number k is given in each test case differently. If there does not exist a k-th password since k is larger than the number of all possible passwords, then you should print `NO' in the output.
Input
Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. The first line of each test case contains one integer, K, the order of the password you should find. Note that 1K
7, 777. Next the following 6 lines show the 6 rows of the first grid and another 6 lines represent the 6 rows of the second grid.
Output
Your program is to write to standard output. Print exactly the k-th password (including ` NO') in one line for each test case.
The following shows sample input and output for three test cases.
Sample Input
3
1
AYGSU
DOMRA
CPFAS
XBODG
WDYPK
PRXWO
CBOPT
DOSBG
GTRAR
APMMS
WSXNU
EFGHI
5
AYGSU
DOMRA
CPFAS
XBODG
WDYPK
PRXWO
CBOPT
DOSBG
GTRAR
APMMS
WSXNU
EFGHI
64
FGHIJ
EFGHI
DEFGH
CDEFG
BCDEF
ABCDE
WBXDY
UWYXZ
XXZFG
YYFYH
EZWZI
ZGHIJ
Sample Output
ABGAG
ABGPS
NO
思路:
解码与编码问题。去掉不重复的字母得出每个位置可能的字母集,sort后递归求解。注意
K-=k*m
string 与char 有别,编程时需要注意。
代码:
#include<iostream>
#include<string>
#include<vector>
#include<cstdio>
#include<algorithm> //sort
#define FOR(a,b,c) for(int a=(b);a<(c);a++)
using namespace std; const int maxn = ,maxm=; int K;
vector<char> table[maxm];
string ans; int dfs(int d) {
if(d==maxm) return true;
int m=; FOR(i,d+,maxm) m *= table[i].size();
FOR(k,,table[d].size())
if(k*m < K && K<=(k+)*m ){
ans += table[d][k];
K-=k*m;
return dfs(d+);
}
return false;
} int main() {
string A[maxn],B[maxn];
int T; scanf("%d",&T);
while(T--) {
ans=""; FOR(i,,maxm) table[i].clear(); //clear() scanf("%d",&K);
FOR(i,,maxn) cin>>A[i];
FOR(i,,maxn) cin>>B[i];
FOR(j,,maxm)
FOR(i,,maxn) FOR(k,,maxn) //make_table
if(A[i][j]==B[k][j]) {
table[j].push_back(A[i][j]);
break;
}
FOR(i,,maxm) sort(table[i].begin(),table[i].end()); //sort
/* FOR(i,0,maxm){
FOR(j,0,table[i].size())
cout<<table[i][j];
cout<<endl;
}
*/
if( dfs() ) cout<<ans; else cout<<"NO";
cout<<"\n";
}
return ;
}
【暑假】[数学]UVa 1262 Password的更多相关文章
- UVA 1262 Password 暴力枚举
Password Time Limit: 3000ms Memory Limit: 131072KB This problem will be judged on UVA. Original ID: ...
- UVa 1262 - Password(解码)
链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem& ...
- UVA 1262 Password
https://vjudge.net/problem/UVA-1262 字典序第k小 注意两点: 1. k-- 2.去重 #include<cstring> #include<cst ...
- UVA - 1262 Password(密码)(暴力枚举)
题意:给两个6行5列的字母矩阵,找出满足如下条件的“密码”:密码中的每个字母在两个矩阵的对应列中均出现.给定k(1<=k<=7777),你的任务是找出字典序第k小的密码.如果不存在,输出N ...
- UVA - 1262 数学
UVA - 1262 题意: 有两个6*5 的大写字母组成的矩阵,需要找出满足条件的字典序第k小的密码:密码中每个字母在两个矩阵的对应的同一列中都出现过 代码: // 先处理出来每一列可以取的字母,例 ...
- 【暑假】[数学]UVa 10375 Choose and divide
UVa 10375 Choose and divide 题目: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=19601 思路 ...
- UVa 1262 (第k字典序) Password
题意: 给出两个6行5列的字母矩阵,一个密码满足:密码的第i个字母在两个字母矩阵的第i列均出现. 然后找出字典序为k的密码,如果不存在输出NO 分析: 我们先统计分别在每一列均在两个矩阵出现的字母,然 ...
- UVa 902 - Password Search
题目:给你一个小写字母组成大的串和一个整数n.找到里面长度为n出现最频繁的子串. 分析:字符串.hash表.字典树. 这里使用hash函数求解,仅仅做一次扫描就可以. 说明:假设频率同样输出字典序最小 ...
- 紫书 例题 10-8 UVa 1262 (暴力枚举)
递归一遍遍历所有情况就ok了 #include<cstdio> #include<cstring> #define REP(i, a, b) for(int i = (a); ...
随机推荐
- js实现网页图片延时加载的原理和代码 提高网站打开速度
有时我们看到一些大型网站,页面如果有很多图片的时候,当你滚动到相应的行时,当前行的图片才即时加载的,这样子的话页面在打开只加可视区域的图片,而其它隐藏的图片则不加载,一定程序上加快了页面加载的速度,对 ...
- 阿里云,CentOS下yum安装mysql,jdk,tomcat
首先说明,服务器是阿里云的,centos6.3_64位安全加固版.首先需要登陆进来,使用的是putty,因为最初的时候,Xshell登陆会被拒绝. 0. 创建个人文件夹 # 使用 yum 安装tomc ...
- HDU4548+素数
简单题. /* */ #include<stdio.h> #include<string.h> #include<stdlib.h> #include<alg ...
- PowerDesigner将name自动添加到Comment注释的方法 VB代码
Option Explicit ValidationMode = True InteractiveMode = im_Batch Dim mdl ' the current model ' get t ...
- Web Api 入门
http://www.cnblogs.com/developersupport/p/WebAPI-Security.html http://www.cnblogs.com/r01cn/archive/ ...
- 网上图书商城项目学习笔记-037工具类之BaseServlet及统一中文编码
1.统一中文编码分析 tomcat默认esetISO-8859-1编码,在servlet中,可能通过request的setCharacterEncoding(charset)和response.set ...
- 使用QGridLayout布局实现翻页效果
http://blog.csdn.net/u013704336/article/details/51474942
- SQL 2008 R2 启动失败 提示 请求失败或服务未及时响应
为什么启动sql server 配置管理器出现请求失败或服务未及时响应_百度知道 http://zhidao.baidu.com/link?url=ElemzIan6I2CqJsd7-7uk5TV25 ...
- editplus的配置文件来支持sql语法高亮【转】
editplus默认是没有sql语法高亮的,原因是它的内部没有sql.stx的这样一个语法文件 我们自己在 EditPlus 的安装目录下面新建一个文件名为sql.stx,然后打开editplus ...
- 一步一步制作yaffs/yaffs2根文件系统(二)---安装BusyBox,构造/bin、/sbin、/usr、linuxr
开发环境:Ubuntu 12.04 开发板:mini2440 256M NandFlash 64M SDRAM 交叉编译器:arm-linux-gcc 4.4.3点此可下载 BusyBox版本: ...