1035 Password (20 分)

最新推荐文章于 2022-08-18 16:15:10 发布

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最新推荐文章于 2022-08-18 16:15:10 发布

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原文链接：http://www.cnblogs.com/zllwxm123/p/11087153.html

本文介绍了一个编程挑战，旨在解决PAT竞赛中密码混淆的问题。通过将容易混淆的字符如1（one）、0（zero）、l和O替换为@、%、L和o，以增强密码的可读性和区分度。文章提供了输入输出规范，并附带了C++代码实现，展示了如何读取用户账户和密码，检查并修改混淆的密码。

1035 Password (20 分)

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

Output Specification:

For each test case, first print the number

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


水题不多说

 1 #include <bits/stdc++.h>
 2 #define N 1050
 3 using namespace std;
 4 int n;
 5 struct Node{
 6     string st, ss;
 7     bool flag;
 8 }node[N];
 9 string st, ss;
10 int main(){
11     int sum = 0;
12     cin >> n;
13     for(int i = 0; i < n; i++){
14         cin >> st >> ss;
15         bool prime = false;
16         for(int i = 0; i < ss.length(); i++){
17             if(ss[i] == '1'){
18                 prime = true;
19                 ss[i] = '@';
20             }else if(ss[i] == '0'){
21                 prime = true;
22                 ss[i] = '%';
23             }else if(ss[i] == 'l'){
24                 prime = true;
25                 ss[i] = 'L';
26             }else if(ss[i] == 'O'){
27                 prime = true;
28                 ss[i] = 'o';
29             }
30         }
31         if(prime){
32             node[sum].st = st;
33             node[sum++].ss = ss;
34         }
35     }
36     if(sum == 0){
37         if(n == 1){
38             cout << "There is 1 account and no account is modified"<<endl;
39         }else{
40             cout << "There are "<<n<<" accounts and no account is modified"<<endl;
41         }
42     }else{
43         cout << sum << endl;
44         for(int i = 0; i < sum; i++)
45             cout << node[i].st << " " << node[i].ss<<endl;
46     }
47     return 0;
48 }