PAT甲级1035

1035 Password (20 point(s))

  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

题目大意

  进行PAT考试的时候，系统会为每个用户随机生成密码，但是有些字符看起来容易混淆，所以用@代替1，%代替0，L代替l，o代替O，现给出N个用户名及其密码，逐个判断密码是否需要修改，最后输出需要修改的密码总数，并输出每个需要修改的用户名及其密码

解题思路

  首先建立一个map，对应需要修改的字符及修改后的字符再逐个判断是否需要修改，若需要，则把用户名及修改后的密码添加到vector中，最后输出vector中的个数及其内容即可

代码

#include 
#include 
#include 
#include 
using namespace std;
map<char, char> mod{ {'1','@'},{'0' , '%'},{'l' , 'L'},{'O' , 'o'} };
int main()
{
	int N, i;
	string id, pwd;
	cin >> N;
	vector<string> res;
	for (i = 0; i < N; ++i) {
		int flag = 0;
		cin >> id >> pwd;
		for (size_t j = 0; j < pwd.size(); ++j) {
			if (mod.find(pwd[j]) != mod.end()) {
				flag = 1;
				pwd[j] = mod[pwd[j]];
			}			
		}
		if (flag) {
			res.push_back(id + " " + pwd);
		}
	}
	if (res.empty()) {
		if (N > 1) printf("There are %d accounts and no account is modified", N);
		else printf("There is 1 account and no account is modified");
	}
	else {
		printf("%d\n", res.size());
		for (i = 0; i < (int)res.size(); ++i)
			printf("%s\n", res[i].c_str());
	}
	return 0;
}