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 (≤), followed by Nlines 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 <iostream>
#include <string>
#include <vector>
using namespace std;
int N;
int main()
{
cin >> N;
vector<pair<string, string>>res;
for (int i = ; i < N; ++i)
{
bool flag = false;
string name, password;
cin >> name >> password;
for (int j = ; j < password.length(); ++j)
{
if (password[j] == '')
{
flag = true;
password[j] = '@';
}
else if (password[j] == '')
{
flag = true;
password[j] = '%';
}
else if (password[j] == 'l')
{
flag = true;
password[j] = 'L';
}
else if (password[j] == 'O')
{
flag = true;
password[j] = 'o';
}
}
if (flag)
res.push_back(make_pair(name, password));
}
if (res.size() == )
{
if (N > )
cout << "There are " << N << " accounts and no account is modified" << endl;
else
cout << "There is " << N << " account and no account is modified" << endl;
}
else
{
cout << res.size() << endl;
for (auto ptr : res)
cout << ptr.first << " " << ptr.second << endl;
}
return ;
}

PAT甲级——A1035 Password的更多相关文章

  1. PAT 甲级 1035 Password (20 分)

    1035 Password (20 分) To prepare for PAT, the judge sometimes has to generate random passwords for th ...

  2. PAT 甲级 1035 Password

    https://pintia.cn/problem-sets/994805342720868352/problems/994805454989803520 To prepare for PAT, th ...

  3. 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 ...

  4. PAT甲级——1035 Password (20分)

    To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem ...

  5. PAT甲级题解(慢慢刷中)

    博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给 ...

  6. A1035 Password (20)(20 分)

    A1035 Password (20)(20 分) To prepare for PAT, the judge sometimes has to generate random passwords f ...

  7. PAT甲级代码仓库

    大道至简,知易行难.希望能够坚持刷题. PAT甲级真题题库,附上我的代码. Label Title Score Code Level 1001 A+B Format 20 1001 * 1002 A+ ...

  8. PAT甲级1131. Subway Map

    PAT甲级1131. Subway Map 题意: 在大城市,地铁系统对访客总是看起来很复杂.给你一些感觉,下图显示了北京地铁的地图.现在你应该帮助人们掌握你的电脑技能!鉴于您的用户的起始位置,您的任 ...

  9. PAT甲级1127. ZigZagging on a Tree

    PAT甲级1127. ZigZagging on a Tree 题意: 假设二叉树中的所有键都是不同的正整数.一个唯一的二叉树可以通过给定的一对后序和顺序遍历序列来确定.这是一个简单的标准程序,可以按 ...

随机推荐

  1. 阿里云CentOs7上安装GitLab

    一.安装 基本上可以根据官网的教程来安装:https://www.gitlab.com.cn/installation/#centos-7 只不过我们暂时没有邮件服务器,所以postfix没有安装. ...

  2. DXP 笔记

    1. 从原理图上添加 net class,快捷键 : P -> V -> C

  3. 7_2.springboot2.x启动配置原理_2.运行run方法

    当创建完SpringApplication对象之后运行run方法 public ConfigurableApplicationContext run(String... args) { StopWat ...

  4. ztree 树状图——例

    效果: <!DOCTYPE html> <html> <head> <meta charset="utf-8"> <meta ...

  5. opencv编译:opencv 3.4.1 编译 contrib模块,增加人脸识别

    start cmake-gui select the opencv source code folder and the folder where binaries will be built (th ...

  6. 【模板篇】splay(填坑)+模板题(普通平衡树)

    划着划着水一不小心NOIP还考的凑合了… 所以退役的打算要稍微搁置一下了… 要准备准备省选了…. 但是自己已经啥也不会了… 所以只能重新拾起来… 从splay开始吧… splay我以前扔了个板子来着, ...

  7. C++:多线程002

    https://blog.csdn.net/morewindows/article/details/7442333 程序描述:主线程启动10个子线程并将表示子线程序号的变量地址作为参数传递给子线程.子 ...

  8. 小米手机 DELETE_FAILED_INTERNAL_ERROR Error while Installing APKs

    手机:小米2s,MIUI 9 7.11.16 开发版 手机已处于开发者模式,启用了USB调试,已使用USB线连接了手机,在Android Studio 工具栏点击 "Run ‘app’(Sh ...

  9. [转]关于 MVVMLight 设计模式系列

    关于 MVVMLight 设计模式系列 http://www.wxzzz.com/958.html MvvmLight框架使用入门 http://www.cnblogs.com/manupstairs ...

  10. 洛谷P1640 【SCOI2010】连续攻击游戏

    原题传送门 题目描述 lxhgww最近迷上了一款游戏,在游戏里,他拥有很多的装备,每种装备都有2个属性,这些属性的值用[1,10000]之间的数表示.当他使用某种装备时,他只能使用该装备的某一个属性. ...