PAT 1035 Password

该博客讨论了PAT比赛前,为用户生成随机密码时遇到的问题,即数字1、0与字母l、O易混淆。为解决此问题,提出了将1替换为@,0替换为%,l替换为L,O替换为o的方案。文章提供了输入输出示例,并给出了相应的代码实现,用于检查和修改可能引起混淆的密码。

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问题描述

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

思路分析及代码实现

更改密码,把给定的元素替换掉,输出更改后的密码

n = int(input())
li = {'1': '@', '0': '%', 'l': 'L', 'O': 'o'}
res = []
jue = 0
for m in range(n):
    a = input().split(' ')
    for p in a[1]:
        if p in li:
            jue = 1
            a[1] = a[1].replace(p, li[p])
    if jue == 1:
        res.append((a[0], a[1]))
        jue = 0
if len(res) != 0:
    print(len(res))
    for j, k in res:
        print(j, k)
else:
    if n == 1:
        print('There is 1 account and no account is modified')
    else:
        print('There are {} accounts and no account is modified'.format(n))

is 写成了are在那瞅半天。。。
在这里插入图片描述

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