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題目

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

將密碼中的幾個字符轉(zhuǎn)化為另一個字符，很簡單的題目。

我的具體做法是邊讀邊改，并且將無需更改的密碼首字符置為'\0'，就是當(dāng)做標(biāo)記以便區(qū)分。后面輸出就容易了。

代碼

最新代碼@github，歡迎交流

#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;
}