1035. Password (20)

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

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给出若干个帐号和密码，如果密码中有1,l,0,O这些会造成疑惑字符，就分别替换成@,%,L,o，如果有密码被修改过，输出总共有多少个密码被修改且输出被修改过的帐号和密码，如果没有则输出"There is(are) n(n,n>=1) account(s) and no account is modified"，注意n为1时前面为there is，account不用加s；n>1时前面为there are，account要加s。

代码：

#include <iostream>#include <cstring>#include <vector>#include <cstdlib>#include <cstdio>using namespace std;bool ismod(string &s){bool flg=false;for(int i=0;i<s.size();i++){switch(s[i]){case '1':s[i]='@';flg=true;break;case '0':s[i]='%';flg=true;break;case 'l':s[i]='L';flg=true;break;case 'O':s[i]='o';flg=true;break;default:break;}}return flg;}int main(){int n;cin>>n;vector<string>ids;vector<string>pws;for(int i=0;i<n;i++){string id,pw;cin>>id>>pw;if(ismod(pw)){ids.push_back(id);pws.push_back(pw);}}int m=ids.size();if(m==0){printf("There %s %d account%s and no account is modified",(n==1)?"is":"are",n,(n==1)?"":"s");}else{cout<<m<<endl;for(int i=0;i<m;i++){cout<<ids[i]<<" "<<pws[i]<<endl;}}}