PAT甲级 -- 1035 Password (20 分)

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为准备PAT，法官需为用户生成随机密码，但存在混淆问题。解决方案是替换特定字符。需编写程序检查并修改混淆密码，还给出了输入输出规范及示例，最后提到一处代码错误警醒。

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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

#include <iostream>
#include <cstring>
#include <vector>
using namespace std;

struct people
{
    char name[15], pass[15];
}p;

int n;  //记录个数
int modify_n = 0; //记录改变的个数
vector<people> v;  //记录修改过的

bool check(char c[])
{
    bool flag = false;
    int len = strlen(c);
    for(int i = 0; i < len; i++)
    {
        switch (c[i])
        {
            case '1':
                {
                    c[i] = '@';
                    flag = true;
                    break;
                }
            case '0':
                {
                    c[i] = '%';
                    flag = true;
                    break;
                }
            case 'l':
                {
                    c[i] = 'L';
                    flag = true;
                    break;
                }
            case 'O':
                {
                    c[i] = 'o';
                    flag = true;
                    break;
                }
        }

    }

    if(flag)  return true;
    else return false;
}

int main()
{
    scanf("%d", &n);
    int num = n;
    while(num--)
    {
        scanf("%s %s", p.name, p.pass);
        if(check(p.pass))
        {
            modify_n += 1;
            v.push_back(p);
        }
    }

    if(modify_n == 0)
    {
        if(n!=1)
        {
            printf("There are %d accounts and no account is modified", n);
        }else
            printf("There is 1 account and no account is modified");
    }else
    {
        printf("%d\n", modify_n);
        for(int i = 0; i < v.size(); i++)
        {
            printf("%s %s\n", v[i].name, v[i].pass);
        }
    }


    return 0;
}

总结：在num--处，之前写成了n--，这会导致只有1个时出错，警醒！！