PAT 1035 Password [字符串处理]

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本文介绍了一个程序设计任务，该任务旨在解决用户密码中易混淆字符的问题，通过将特定字符替换为不易混淆的符号，确保密码的清晰性和独特性。程序接收一系列账户及其密码作为输入，检查并修正包含易混淆字符的密码，最后输出被修正的账户信息。

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 modifiedwhere 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

我的代码：

#include<stdio.h>
#include<string.h>

//我也不知道我脑回路是怎样的九曲十八弯，明明能原地修改的事儿我非得另外开个数组存取，还因此有了一组错误答案 

int main()
{
	int n,i,j,count = 0,s[1005]={};
	char user[1005][20],psw[1005][20];
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		scanf("%s%s",&user[i],&psw[i]);
		for(j=0;j<strlen(psw[i]);j++)
		{
			if(psw[i][j]=='1') {psw[i][j] = '@';s[i] = 1;}
			else if(psw[i][j]=='0') {psw[i][j] = '%';s[i] = 1;}
		    else if(psw[i][j]=='l') {psw[i][j] = 'L';s[i] = 1;}
		    else if(psw[i][j]=='O') {psw[i][j] = 'o';s[i] = 1;}
		}
		//s[i]的1/0来存取每组修改状态以便最后输出的判断			
		if(s[i])  count++;	
	}
	if(count==0)
	{
		if(n==1) printf("There is 1 account and no account is modified\n");
		else printf("There are %d accounts and no account is modified\n",n);
	}
	else
	{
	    printf("%d\n",count);
		for(i=0;i<n;i++)
		    if(s[i])  printf("%s %s\n",user[i],psw[i]);
	}
}