pat1035. Password (20)
1035. Password (20)
To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem is that there are always some confusing passwords since it is hard to distinguish 1 (one) from l (L in lowercase), or 0 (zero) from O (o in uppercase). One solution is to replace 1 (one) by @, 0 (zero) by %, l by L, and O by o. Now it is your job to write a program to check the accounts generated by the judge, and to help the juge modify the confusing passwords.
Input Specification:
Each input file contains one test case. Each case contains a positive integer N (<= 1000), followed by N lines of accounts. Each account consists of a user name and a password, both are strings of no more than 10 characters with no space.
Output Specification:
For each test case, first print the number M of accounts that have been modified, then print in the following M lines the modified accounts info, that is, the user names and the corresponding modified passwords. The accounts must be printed in the same order as they are read in. If no account is modified, print in one line "There are N accounts and no account is modified" where N is the total number of accounts. However, if N is one, you must print "There is 1 account and no account is modified" instead.
Sample Input 1:
3
Team000002 Rlsp0dfa
Team000003 perfectpwd
Team000001 R1spOdfa
Sample Output 1:
2
Team000002 RLsp%dfa
Team000001 R@spodfa
Sample Input 2:
1
team110 abcdefg332
Sample Output 2:
There is 1 account and no account is modified
Sample Input 3:
2
team110 abcdefg222
team220 abcdefg333
Sample Output 3:
There are 2 accounts and no account is modified
#include<cstdio>
#include<stack>
#include<algorithm>
#include<iostream>
#include<stack>
#include<set>
#include<map>
#include<vector>
using namespace std;
vector<string> v;
map<string,string> ha;
bool check(char &c){
if(c==''){
c='@';
return true;
}
if(c==''){
c='%';
return true;
}
if(c=='l'){
c='L';
return true;
}
if(c=='O'){
c='o';
return true;
}
return false;
}
int main(){
//freopen("D:\\INPUT.txt","r",stdin);
int n;
scanf("%d",&n);
int i,j;
string num,pas;
int count=;
for(i=;i<n;i++){
cin>>num>>pas;
bool can=false;
for(j=;j<pas.length();j++){
if(check(pas[j])){
can=true;
}
}
if(can){
count++;
v.push_back(num);
ha[num]=pas;
}
}
if(count){
printf("%d\n",count);
for(j=;j<v.size();j++){
cout<<v[j]<<" "<<ha[v[j]]<<endl;
}
}
else{
if(n==){
printf("There is 1 account and no account is modified\n",n);
}
else{
printf("There are %d accounts and no account is modified\n",n);
}
}
return ;
}
pat1035. Password (20)的更多相关文章
- PAT1035: Password
1035. Password (20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue To prepare f ...
- A1035 Password (20)(20 分)
A1035 Password (20)(20 分) To prepare for PAT, the judge sometimes has to generate random passwords f ...
- PAT 甲级 1035 Password (20 分)(简单题)
1035 Password (20 分) To prepare for PAT, the judge sometimes has to generate random passwords for ...
- PAT (Advanced Level) Practice 1035 Password (20 分) 凌宸1642
PAT (Advanced Level) Practice 1035 Password (20 分) 凌宸1642 题目描述: To prepare for PAT, the judge someti ...
- PAT甲级——1035 Password (20分)
To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem ...
- 1035 Password (20)
#include <stdio.h> #include <string.h> struct MyStruct { ]; ]; bool changed; }; int main ...
- 【PAT】1035. Password (20)
题目:http://pat.zju.edu.cn/contests/pat-a-practise/1035 分析:简单题.直接搜索,然后替换,不会超时,但是应该有更好的办法. 题目描述: To pre ...
- 1035 Password (20)(20 point(s))
problem To prepare for PAT, the judge sometimes has to generate random passwords for the users. The ...
- PAT A1035 Password (20)
AC代码 注意创造函数条件中使用引用 输出语句注意单复数 #include <cstdio> #include <cstring> #include <iostream& ...
随机推荐
- js函数篇
1.闭包函数,作用:不污染全局变量, 定义:与外界隔离的独立作用域被称为闭包,使用函数实现该功能称为函数闭包: 写法: (function(){ function sayHello(){ conso ...
- 关于 vs 2012 键盘无法输入的问题
使用vs2012 新建了一个类文件之后,vs2012的编辑界面突然出现奇怪的问题,键盘无法输入! 最后调查的结果是由于resharper插件导致的. 可以将插件禁用然后启用. 也可以删除resharp ...
- error: converting to execution character set: Invalid or incomplete multibyte or wide character
交叉编译.c文件,遇到如下问题 arm-linux-gcc -o show_lines show_lines.c -lfreetype -lm show_lines.c:199:19: error: ...
- net start sql server (instance)
如何启动 SQL Server 实例(net 命令) 其他版本 可以使用 Microsoft Windows net 命令启动 Microsoft SQL Server 服务. 启动 SQL Se ...
- zabbix 系列 (1)安装
安装server http://blog.csdn.net/xiegh2014/article/details/54988548 安装 agent http://m.blog.csdn.net/wu2 ...
- MS-SQL使用xp_cmdshell命令导出数据到excel
exec master..xp_cmdshell 'bcp "select c.Category_Title as 标题,p.Category_Title as 所属分类 from ltbl ...
- day01_虚拟机与主机之间ip配置
虚拟机1: centos_ node1 虚拟机2:centos_node2 宿主主机虚拟机ip配置: vmnet1 来自为知笔记(Wiz)
- [hdu1176]免费馅饼(数塔dp)
题意:中文题,不解释了 = = 解题关键:逆推,转化为数塔dp就可以了 dp[i][j]表示在i秒j位置的最大值. 转移方程:$dp[i][j] = \max (dp[i + 1][j],dp[i + ...
- Hadoop添加节点datanode(生产环境)
Hadoop添加节点datanode 博客分类: hadoop HadoopSSHJDKXML工作 1.部署hadoop 和普通的datanode一样.安装jdk,ssh 2.修改host ...
- 6.6 chmod的使用
从公司拷贝了白天整理的笔记,拿回家整理,结果发现有锁,无法对其解压.解决方案如上: ll 命令,查看其权限. sudo chmod 777 Picture.tar-1修改权限. 然后,可以正常打开Pc ...