PAT A1035 Password

1035 Password

分数 20

作者 CHEN, Yue

单位 浙江大学

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

* 将密码含有 1 ,l, 0, O的字符换为对应的字符，然后输出修改后的密码；
 *
 * 修改字符可以用一个map来存储，如果有修改的密码可以用vector来存储，最后判断vector的
 * 值，根据题目要求输出相应的内容。

/**
 * 将密码含有 1 ,l, 0, O的字符换为对应的字符，然后输出修改后的密码；
 * 
 * 修改字符可以用一个map来存储，如果有修改的密码可以用vector来存储，最后判断vector的
 * 值，根据题目要求输出相应的内容。
*/

#include <iostream>
#include <map>
#include <string>
#include <vector>

using namespace std;

typedef pair<string, string> PSS;

map<char, char> mp = {{'1', '@'}, {'0', '%'}, {'l', 'L'}, {'O', 'o'}};

string modify(string word)
{
    string s;
    for(auto ele : word)
    {
        if(mp.find(ele) != mp.end())
            ele = mp[ele];
        s.push_back(ele);
    }
    
    return s;
}

int main()
{
    int n;
    cin >> n;
    
    vector<PSS> res;
    
    for(int i=0; i<n; ++i)
    {
        string name, word;
        cin >> name >> word;
        string s = modify(word);
        if(s != word)   res.push_back({name, s});
    }
    
    if(res.size() == 0)
    {
        if(n == 1)
            puts("There is 1 account and no account is modified");
        else 
            printf("There are %d accounts and no account is modified\n", n);
    }
    else
    {
        cout << res.size() << endl;
        for(auto ele : res)
            cout << ele.first << ' ' << ele.second << endl;
    }
    
    return 0;
}