算法精髓

利用c++特性中的引用机制，我的理解是：引用 == 给地址起别名，本质上还是传地址。

通过引用，可以直接在另一个函数进行判断和修改，然后在主函数中使用。

具体如下


struct node {
    char name[12], password[12];
    bool ischange;
}T[1005];

void trans(node& t, int& rep)
{
    int len = strlen(t.password);
    for (int i = 0; i < len; i++) {
        if (t.password[i] == '1') {
            t.password[i] = '@';
            t.ischange = true;
        } else if (t.password[i] == '0') {
            t.password[i] = '%';
            t.ischange = true;
        } else if (t.password[i] == 'l') {
            t.password[i] = 'L';
            t.ischange = true;
        } else if (t.password[i] == 'O') {
            t.password[i] = 'o';
            t.ischange = true;
        }
    }
    if (t.ischange)
        rep++;
}

注解

关于这道题的解决，留意最后的条件输出就好，关于英文单复数的。

原题

1035 Password (20 point(s))

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

 

 

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