1035 Password (20 分)

密码混淆与修正

最新推荐文章于 2024-05-21 17:00:05 发布

haohao0626

最新推荐文章于 2024-05-21 17:00:05 发布

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分类专栏： PAT 甲级

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本文介绍了一个程序设计问题，旨在解决由于数字和字母相似性引起的密码混淆。通过将易混淆的字符如1、0、l、O替换为@、%、L、o，从而增强密码的可读性和区分度。文章提供了代码示例，展示了如何实现这一功能。

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

细节：

#include <iostream>
#include <string>
#include <vector>	
#include <queue>

using namespace std;

struct password
{
	string id, pw;
};



int modify(string &s)
{
	int count = 0;
	for (auto &c : s)
	{
		switch (c)
		{
		default:
			count++;  //如果count的值和s的长度相同  即没有改变其中的某个字符
			break;
		case '1':
			c = '@';
			break;
		case '0':
			c = '%';
			break;
		case 'l':
			c = toupper(c);
			break;
		case 'O':
			c = tolower(c);
			break;
		}
	}
	return count;
}

int main()
{
	//freopen("in.txt", "r", stdin);
	int n, finalcount = 0;
	cin >> n;
	vector<password> p(n);
	queue<int> m;  //用来统计哪个字符串改了   输出的时候得知道输出哪个
	for (int i = 0; i < n; i++)
	{
		cin >> p[i].id >> p[i].pw;
	}
	for (int i = 0; i < n; i++)
	{
		if (modify(p[i].pw) != p[i].pw.length())
		{
			m.push(i); //第几个改了  把它的下标存起来
			finalcount++;
		}
	}
	if (!finalcount) //没有改
	{
		
		/*这是最先想到的写法    这种其实就是考验你的细节  细心
		if (n == 1)   //n=1输出写“1 account”  不是1的话 得加s  “accounts” 还有 is和are的区别 坑爹啊  看柳婼代码 迅速意识到了这个细节
		{
			cout << "There is 1 account and no account is modified" << endl;
		}
		else
		{
			cout << "There are" << " " << n << " accounts and no account is modified" << endl;
		}*/



		//第二种写法   从c++ primer 上看到的
		cout << "There " << (n == 1 ? "is " : "are ") << n
			<< (n == 1 ? " account" : " accounts") << " and no account is modified" << endl;
	}
	else
	{
		cout << finalcount << endl;
		while (!m.empty())
		{
			cout << p[m.front()].id << " " << p[m.front()].pw << endl;
			m.pop();
		}
	}
	//fclose(stdin);
	return 0;
}