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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 R1spOdfaSample Output 1:
2 Team000002 RLsp%dfa Team000001 R@spodfaSample Input 2:
1 team110 abcdefg332Sample Output 2:
There is 1 account and no account is modifiedSample Input 3:
2 team110 abcdefg222 team220 abcdefg333Sample Output 3:
There are 2 accounts and no account is modified
#pragma warning(disable:4996)#include <stdio.h>#include <iostream>#include <string>#include <vector>using namespace std;struct Team {string name;string password;bool isModified = false;};vector<Team> t;int main(void) {int n;cin >> n;for (int i = 0; i < n; i++) {Team temp;cin >> temp.name >> temp.password;t.push_back(temp);}int count=0;for (int i = 0; i < n; i++) {string result=t[i].password;for (int j = 0; j < result.length(); j++) {if (result[j] == ‘1‘)result[j] = ‘@‘;if (result[j] == ‘0‘)result[j] = ‘%‘;if (result[j] == ‘l‘)result[j] = ‘L‘;if (result[j] == ‘O‘)result[j] = ‘o‘;}if (result != t[i].password) {count++;t[i].isModified = true;t[i].password = result;}}if (count != 0) {cout << count << endl;for (int i = 0; i < n; i++) {if (t[i].isModified == true) {cout << t[i].name << " " << t[i].password<<endl;}}}else if (n == 1) {cout << "There is 1 account and no account is modified";}elsecout << "There are "<<n<<" accounts and no account is modified";return 0;}
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原文地址：http://www.cnblogs.com/zzandliz/p/5023129.html
