PAT Advanced 1035 Password (20分)

Posted 2020-11-25 littlepage

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 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

这道题考察了字符替换和逻辑，乙级有相同的题目。

这边要注意的是输出时，单数是account，复数是accounts

#include <iostream>
#include <map>
using namespace std;
int main() {
    int N;
    cin >> N;
    string user, pw, res;
    map<char, char> m;
    m[‘l‘] = ‘L‘;m[‘O‘] = ‘o‘;
    m[‘1‘] = ‘@‘;m[‘0‘] = ‘%‘;
    int modicoun = 0;
    for(int j = 0; j < N; j++) {
        bool modi = false;
        cin >> user >> pw;
        for(int i = 0; i < pw.length(); i++){
            if(m[pw[i]] != 0) {
                pw[i] = m[pw[i]];
                modi = true;
            }
        }
        if(modi) {
            res = res + user + " " + pw + "
";
            modicoun++;
        }
    }
    if(!modicoun) {
        if(N != 1) printf("There are %d accounts and no account is modified", N);
        else printf("There is 1 account and no account is modified");
    }else {
        printf("%d
%s", modicoun, res.c_str());
    }
    return 0;
}