PAT A1035

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该博客介绍了PAT A1035编程题目，内容涉及检查和修改可能引起混淆的随机密码，如用@替换1，%替换0，L替换l，o替换O。程序需要读取输入文件中的账户信息，输出修正后的账户数量及详情。

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PAT A1035

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;
}

这里写图片描述