PAT (Advanced Level) Practise 1035

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本文介绍了一个用于检查并修改PAT竞赛中可能产生混淆的密码的程序。该程序将1替换为@、0替换为%、l替换为L、O替换为o，确保密码易于区分。文章提供了完整的C语言实现代码，并通过样例展示了输入输出格式。

PAT (Advanced Level) Practise 1035

1035. Password (20)

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

编程实现

#include <stdio.h>
#define N 1000

typedef struct {
    char name[12];
    char passwd[12];
    int modify;
} Account;

int modify(char *info)
{
    int flag = 0;
    char *c = info;

    while ( *c != '\0' ) {
        if ( *c == '1' ) {
            *c = '@';
            flag = 1;
        } else if ( *c == '0' ) {
            *c = '%';
            flag = 1;
        } else if ( *c == 'l' ) {
            *c = 'L';
            flag = 1;
        } else if ( *c == 'O' ) {
            *c = 'o';
            flag = 1;
        }
        c++;
    }
    return flag;
}

int main(void)
{
    Account accounts[N];
    int i, n, m = 0;
    int change = 0;

    scanf("%d", &n);
    for ( i = 0; i < n; i++ ) {
        scanf("%s %s", accounts[i].name, 
                accounts[i].passwd);
        change = modify(accounts[i].passwd);
        accounts[i].modify = change;
        m += change;
    }
    if ( m == 0 && n == 1 ) 
        printf("There is 1 account and no account is modified\n");
    else if ( m != 0 ) {
        printf("%d\n", m);
        for ( i = 0; i < n; i++ ) {
            if ( accounts[i].modify ) 
                printf("%s %s\n", accounts[i].name, 
                    accounts[i].passwd);
        }
    } else
        printf("There are %d accounts and no account is modified\n", n);

    return 0;
}