PAT甲级 1035. Password (20)

题目：

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:

3Team000002 Rlsp0dfaTeam000003 perfectpwdTeam000001 R1spOdfa

Sample Output 1:

2Team000002 RLsp%dfaTeam000001 R@spodfa

Sample Input 2:

1team110 abcdefg332

Sample Output 2:

There is 1 account and no account is modified

Sample Input 3:

2team110 abcdefg222team220 abcdefg333

Sample Output 3:

There are 2 accounts and no account is modified

思路：

这道题很简单，只是简单地将字符串中特定的字符进行替换

代码：

#include<iostream>#include<string>#include<vector>using namespace std;struct user{string name;string password;};int main(){

//输入int N;cin >> N;vector<user> users(N);int i = 0;for (; i < N; ++i){cin >> users[i].name >> users[i].password;}//替换int j;vector<user> modified;bool flag = 0;for (i = 0; i < N; ++i){flag = 0;for (j = 0; j < users[i].password.size(); ++j){switch (users[i].password[j]){case '1':{flag = 1;users[i].password[j] = '@';break;}case '0':{flag = 1;users[i].password[j] = '%';break;}case 'l':{flag = 1;users[i].password[j] = 'L';break;}case 'O':{flag = 1;users[i].password[j] = 'o';break;}}}if (flag){modified.push_back(users[i]);}}//输出if (modified.size() == 0){if (users.size() == 1)cout << "There is 1 account and no account is modified" << endl;elsecout << "There are " << N << " accounts and no account is modified" << endl;}else{cout << modified.size() << endl;for (i = 0; i < modified.size(); ++i){cout << modified[i].name << " " << modified[i].password << endl;}}system("pause");return 0;}