PAT甲级(Advanced Level)练习题 Password(20)

题目描述

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.

输入描述:

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.

输出描述:

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.

输入例子:

3
Team000002 Rlsp0dfa
Team000003 perfectpwd
Team000001 R1spOdfa

输出例子:

2
Team000002 RLsp%dfa
Team000001 R@spodfa

题目分析及实现

本体的目的是要把字符串中对应的几个字符进行修改。难度比较低，只需要顺序查找每个字符，然后用switch语句进行替换，并把进行了替换的字符的flag设置为1。记录下来，最后可以得到数目，还有修改后的字符串打印出来。

代码实现（C++版本）

#include <iostream>
#include <string>
using namespace std;
struct User{
    string name;
    string pwd;
    int flag = 0;
};

int main()
{
    int num;
    cin >> num;
    User user[num];
    for(int i =0;i<num;i++){
        cin>>user[i].name>>user[i].pwd;
        for(unsigned long j =0;j<user[i].pwd.length();j++){
            switch(user[i].pwd[j]){
            case '0':
                user[i].pwd[j] ='%'; user[i].flag = 1; break;
            case '1':
                user[i].pwd[j] ='@'; user[i].flag = 1; break;
            case 'l':
                user[i].pwd[j] ='L'; user[i].flag = 1; break;
            case 'O':
                user[i].pwd[j] ='o'; user[i].flag = 1; break;
            }
        }
    }
    int count = 0;
    for(int i =0;i<num;i++){
        if(user[i].flag != 0){
            count++;
        }
    }

    if (count == 0){
        if(num>1)
        cout<<"There are "<<num <<" accounts and no account is modified"<<endl;
        else{
            cout<<"There is "<<num <<" account and no account is modified"<<endl;
        }
        return 0;
    }
    cout<<count<<endl;
    for(int i = 0;i<num;i++){
        if(user[i].flag==1)
        cout<<user[i].name<<" "<<user[i].pwd<<endl;
    }

    return 0;
}