PAT (Advanced Level)-1035. Password (20)

PAT (Advanced Level)-1035. Password (20)

1035. Password (20)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue


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

解读

本题目考查点在于对字符串的检索处理,同时也考查对于输入数据的保存。
易忽视的地方: C++中,开数组存储的数组大小,cin缓冲流存在 特殊情况N=0的时候。

另外想吐槽的一句是:这个输出结束之后为什么没有句号啊!!我带了句号为什么还对了三个检查点!!!害我检查问题查了好久才注意到句末不带标点的!!!!

完整代码

#include<iostream>
#include<vector>
#include<cstring>

using namespace std;

typedef struct user
{
	bool legal;
	char Account[11];
	char Password[11];
}User_info;

//determind if the password is legal, and modify it if illegal.
bool Is_legal(user& User_info)
{
	bool legal = true;
	int len = strlen(User_info.Password);

	for (int i = 0; i < len; ++i)
	{
		if (User_info.Password[i] == '1')
		{
			User_info.Password[i] = '@';
			legal = false;
		}
		if (User_info.Password[i] == '0')
		{
			User_info.Password[i] = '%';
			legal = false;
		}
		if (User_info.Password[i] == 'l')
		{
			User_info.Password[i] = 'L';
			legal = false;
		}
		if (User_info.Password[i] == 'O')
		{
			User_info.Password[i] = 'o';
			legal = false;
		}
	}
	return legal;
}

int main()
{
	int N;
	cin >> N;
	vector<User_info> User_infoVec(N);

	int legal = 0;
	for (int i = 0; i < N; ++i)
	{
		cin >> User_infoVec[i].Account;
		cin >> User_infoVec[i].Password;
		if (Is_legal(User_infoVec[i]))
		{
			User_infoVec[i].legal = true;
			legal++;
		}
		else
		{
			User_infoVec[i].legal = false;
		}
	}

	//Output
	if (legal == N)
	{
		if (N <= 1)
		{
			cout << "There is " << N << " account and no account is modified" << endl;
		}
		else
		{
			cout << "There are " << N << " accounts and no account is modified" << endl;
		}
	}
	else
	{
		cout << N - legal << endl;
		for (int i = 0; i < N; ++i)
		{
			if (User_infoVec[i].legal == false)
			{
				cout << User_infoVec[i].Account << " " << User_infoVec[i].Password << endl;
			}
		}
	}
	return 0;
}

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