PAT Advanced 1035. Password (20)

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题目

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
( ), followed by 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 of accounts that have been
modified, then print in the following 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

思路

将密码中的几个字符转化为另一个字符，很简单的题目。

我的具体做法是边读边改，并且将无需更改的密码首字符置为'\0'，就是当做标记以便区分。后面输出就容易了。

代码

最新代码@github，欢迎交流

#include <stdio.h>

int main()
{
    int N, count = 0, modified;
    char username[1000][11], password[1000][11];

    scanf("%d", &N);
    for(int i = 0; i < N; i++)
    {
        modified = 0;
        scanf("%s %s", username[i], password[i]);
        for(char *p = password[i]; *p; p++)
        {
            switch(*p)
            {
                case '1': *p = '@'; modified = 1; break;
                case '0': *p = '%'; modified = 1; break;
                case 'l': *p = 'L'; modified = 1; break;
                case 'O': *p = 'o'; modified = 1; break;
                default: break;
            }
        }
        if(modified)
            count++;
        else  /* mark unmodified password */
            password[i][0] = '\0';
    }

    if(count)
    {
        printf("%d\n", count);
        for(int i = 0; i < N; i++)
            if(password[i][0] != '\0')
                printf("%s %s\n", username[i], password[i]);
    }
    else
    {
        printf("There %s %d %s and no account is modified",
            N > 1 ? "are" : "is", N, N > 1 ? "accounts" : "account");
    }

    return 0;
}