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最新推荐文章于 2024-08-28 08:16:30 发布

Ididcan

最新推荐文章于 2024-08-28 08:16:30 发布

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分类专栏： ACM 文章标签： passwords each integer random user output

本文链接：https://blog.youkuaiyun.com/Ididcan/article/details/4634274

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本文介绍了一个用于处理易混淆密码的程序设计案例。该程序能够自动识别并修正容易引起混淆的字符，如用 '@' 替换 '1'，用 '%' 替换 '0' 等，以提高密码的可读性和区分度。

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Generate Passwords

Time Limit: 1 Second Memory Limit: 32768 KB

To prepare for programming contests or coding examinations, 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

There are multiple test cases. 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. A test case with N = 0 denotes the end of input. This test case is not to be processed.

Output

For each 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 "No account is modified."

Sample Input:

2
Team000001 R1spOdfa
Team000002 Rlsp0dfa
1
team110 abcdefg332
0

Sample Output:

2
Team000001 R@spodfa
Team000002 RLsp%dfa
No account is modified.
//2514
#include <iostream>
using namespace std;

int main()
{
	char u[1000][100],p[1000][100];
	int flag[1000];
	int n;
	while(cin>>n&&n!=0)
	{
		int j = 0;
		int num = 0;
		while(j<n)
		{
			scanf("%s %s",u[j],p[j]);
			int l = strlen(u[j]);
			flag[j] = 0;
			for ( int i = 0 ; i < strlen(p[j]); i++)
			{
				switch(p[j][i])
				{
					case '0':
						flag[j] = 1;p[j][i] = '%';break;
					case '1':
						flag[j] = 1;p[j][i] = '@';break;
					case 'l':
						flag[j] = 1;p[j][i] = 'L';break;
					case 'O':
						flag[j] = 1;p[j][i] = 'o';break;
				}
			}
			if(flag[j]==1)
				num++;
			j++;
		}
		if(num==0)
			printf("No account is modified./n");
		else
		{
			printf("%d/n",num);
			for(int k = 0 ; k < j; k++)
			{
				if(flag[k]==1)
					printf("%s %s/n",u[k],p[k]);
			}
		}
	}
	return 0;
}