1035. Password (20)

题目如下：

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

为了按照输入的顺序输出，我们使用vector来容纳每一条记录，每一条记录包含名称和密码两项，对于每一条输入，我们定义tmp空字符串，从前到后遍历输入的字符串，对于不需要替换的字符，直接拼接到tmp尾部，如果需要替换，则把替换后的拼接到tmp尾部，最后tmp就是新密码，把name和tmp压入vector作为一条记录。

最后通过vector的规模就可以知道是否有被调整的账户，注意按照题目要求N=1和N!=1时的输出有所差异。

#include <iostream>
#include <string>
#include <string.h>
#include <stdio.h>
#include <vector>

//replace 1 (one) by @, 0 (zero) by %, l by L, and O by o.

using namespace std;

struct Record{

    string name;
    string pwd;

    Record(string _n, string _p) : name(_n), pwd(_p) {}

};

int main()
{
    int N;
    cin >> N;
    string name,pwd;
    vector<Record> records;
    for(int i = 0; i < N; i++){
        cin >> name >> pwd;
        string tmp;
        for(int j  = 0; j < pwd.length(); j++){
            char c = pwd[j];
            switch(c){
            case '1':
                tmp.push_back('@');
                break;
            case '0':
                tmp.push_back('%');
                break;
            case 'l':
                tmp.push_back('L');
                break;
            case 'O':
                tmp.push_back('o');
                break;
            default:
                tmp.push_back(c);
                break;
            }
        }
        if(tmp != pwd){
            records.push_back(Record(name,tmp));
        }
    }
    if(records.size() == 0){
        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{
        cout << records.size() << endl;
        for(int i = 0; i < records.size(); i++){
            cout << records[i].name << " " << records[i].pwd << endl;
        }
    }
    return 0;
}