PAT A1035

PAT A1035

1. Password (20)
   时间限制
   400 ms
   内存限制
   65536 kB
   代码长度限制
   16000 B
   判题程序
   Standard
   作者
   CHEN, Yue

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

这道题真的是太费劲了，简直成了细节性的语文题。

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <string>
#include <vector>
#include <cstring>
#include <cmath>
#include <set>
#include <queue>
#include <algorithm> 
#include <bitset>
#include <sstream> 
#include <iomanip>
#include <map>

using namespace std;






int cnt=0;

string user_n[1000+10],user_p[1000+10];

bool vis[1000+10]={false};


bool isvis(string str){

    for(auto x:str){


        if(x=='1' 
            || x=='l' 
            || x=='0'
            || x=='O' )

        {

            return true;
        }   


    }

        return false;

}


string  change(string des){

    string str=des;

    for(int i=0;i<str.length();i++){

        if(str[i]=='1'){
            str[i]='@';

        }else   if(str[i]=='0'){
            str[i]='%';

        }else   if(str[i]=='l'){
            str[i]='L';

        }else   if(str[i]=='O'){
            str[i]='o';

        } 



    }



    return str;


}




int main(){

    int n;
    cin>>n;

    for(int i=0;i<n;i++){
        string name,pwd;
        cin>>name>>pwd;

        user_n[i]= name;
        user_p[i] = pwd;


    }


    for(int i=0;i<n;i++){

        if(isvis(user_p[i])){
            cnt++;

            vis[i]=true;
            user_p[i]=change(user_p[i]);

        }



    }


    if( cnt ==0){

        if(n==1){
            cout<<"There is 1 account and no account is modified"<<endl;
        }
        else{

            cout<<"There are "<<n<<" accounts and no account is modified"<<endl;
        }



    }
    else{

        cout<<cnt<<endl;

        for(int i=0;i<n;i++){

            if( vis[i]){

                cout<<user_n[i]<<" "<<user_p[i]<<endl;



            }

        }




    }







    return 0;
}