PAT (Advanced Level) Practice 1035 Password (20 分) 凌宸1642

2024-03-05 09:30:18

PAT (Advanced Level) Practice 1035 Password (20 分) 凌宸1642

题目描述：

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 judge modify the confusing passwords.

译：为了准备 PAT , 裁判们有时候不得不为用户生成随机的密码。这样的问题就是经常会出现许多混乱的密码，因为很难区分 1(1) 和 l (L 的小写) , 0 (0) 和O (o 的大写)。一种解决办法就是用 @ 代替 1 ， % 代替 0 , L 代替 l , O代替 o。现在你的任务就是写一个程序检查裁判生成的账号，并帮助裁判修改这混乱的密码。

Input Specification (输入说明):

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.

译：每个输入文件包含一个测试用例，每个用例包含一个正整数 N (≤1000), 紧跟着 N 行账号。每个账号包含一个用户名和一个密码，并且都是不超过十个字符并且不含空白符的字符串

Output Specification (输出说明):

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.

译：对于每个测试用例，第一行输出被修改过的账号的数量 M，然后接下来 M 行打印修改过的账号信息，也就是说，用户名对应的修改过的密码。账号的输出顺序必须和输入顺序一致。如果没有账号被修改，在一行中打印 There are N accounts and no account is modified 其中 N 表示账号的总数。但是，如果 N 等于 1的话，你必须输出 There is 1 account and no account is modified 。

Sample Input 1 (样例输入1):

3

Team000002 Rlsp0dfa

Team000003 perfectpwd

Team000001 R1spOdfa

Sample Output 1 (样例输出1):

2

Team000002 RLsp%dfa

Team000001 R@spodfa

Sample Input 2 (样例输入2):

1

team110 abcdefg332

Sample Output 2 (样例输出2):

There is 1 account and no account is modified

Sample Input 3 (样例输入3):

2

team110 abcdefg222

team220 abcdefg333

Sample Output 3 (样例输出3):

There are 2 accounts and no account is modified

The Idea：

设计到字符串，由于需要存储一个 name - passwd 的键值对，本来首选是使用 map 来进行键值对的处理，但是map会自动排序，与要求输出顺序需要喝输入顺序不同相矛盾，故采取了用 vector来存储 pair 类型数据。值得注意的是，题目中明确说明了 n = 1 和 n != 1的时候的情况输出是不同的，特别注意 are 和 is 还有 account 和 accounts 这里有坑（测试点2) .

The Codes:

#include<bits/stdc++.h>

using namespace std ;

string name , passwd1 , passwd2 ;

int n ;

vector<pair<string , string> > ans ; // 利用 pair 捆绑 name 和 passwd

string modify(string s){

	string str = "" ;

	for(int i = 0 ; i < s.size() ; i ++){

		if(s[i] == '1') str += '@' ; // 如果是 1 替换成 @

		else if(s[i] == '0') str += '%' ; // 如果是 0 替换成 %

		else if(s[i] == 'l') str += 'L' ; // 如果是 l 替换成 L

		else if(s[i] == 'O') str += 'o' ; // 如果是 O 替换成 o

		else str += s[i] ; // 其他的直接连接

	}

	return str ;

}

int main(){

	scanf("%d" , &n) ;

	for(int i = 0 ; i < n ; i ++){

		cin >> name >> passwd1 ;

		passwd2 = modify(passwd1) ;

		if(passwd1 != passwd2) ans.push_back(make_pair(name , passwd2)) ;// 修改的加入向量

	}

	if(ans.size() == 0){

		if( n == 1) printf("There is 1 account and no account is modified\n") ;

		else printf("There are %d accounts and no account is modified\n" , n) ;

	}else{

		printf("%d\n" , ans.size()) ;

		for(int i = 0 ; i < ans.size() ; i ++)

			printf("%s %s\n" , (ans[i].first).c_str() , (ans[i].second).c_str()) ;

	}

	return 0;

}

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