PAT 1035 Password （20 分）String的性质 燚

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

题意： 由于某些数字与一些英文字母很相似可能造成一些混乱，现将密码字符串中的 数字 1 替换为@  数字0 替换为% 小写字母l 替换为L 大写字母O 替换为小写o

输入：第一行：输入正整数N 表示要输入的信息个数

          剩下N行：账号 密码

输出：如果有被修改的字符串，则输出被修改信息的个数以及修改过后的内容

          如果所有密码字符串都没有被修改，若输入数量>1则输出：There are N accounts and no account is modified

      否则输出：There is 1 account and no account is modified

思路：逐个检测字符串的每一位，用switch 语句判断并改整需要修改的字符。

知识点：string类型在内存中为连续存储，类似与字符数组，但string 为连续容器它与vector容器很相似，并且可以共用很多操作函数。

#include<iostream>
#include<vector>
#include<string>
using namespace std;
struct Count {
	string name;
	string psw;
};
void input(vector<Count> &count) {
	int n;
	cin >> n;
	count.resize(n);
	for (int i = 0;i < n;i++) {
		cin >> count[i].name >> count[i].psw;
	}
}
void changeThePsw(vector<Count> &count, vector<int> &index) {
	for (int i = 0;i < count.size();i++) {
		bool flag = false;  //判断密码中是否有可修改的字符，若有flag=true
		int j = 0;
        //检测密码字符串的每一位，并判断是否修改
		for (;j < count[i].psw.size();j++) { 
			switch (count[i].psw[j]) {
			case 'l':count[i].psw[j] = 'L';flag = true;break;
			case 'O':count[i].psw[j] = 'o';flag = true;break;
			case '0':count[i].psw[j] = '%';flag = true;break;
			case '1':count[i].psw[j] = '@';flag = true;break;
			}
		}
		if(flag)
		index.push_back(i);//将修改过后的结构体 下标 存入容器中
	}
}
int main() {
	vector<Count> count;
	vector<int>index;
	input(count);
	changeThePsw(count, index);
	if (index.empty() && count.size() > 1) {
		cout << "There are " << count.size() << " accounts and no account is modified" << endl;
	}
	else if (index.empty() && count.size() == 1) {
		cout << "There is 1 account and no account is modified" << endl;
	}
	else {
		cout << index.size() << endl;
		for (auto i = index.begin();i != index.end();i++) {
			cout << count[(*i)].name << " " << count[(*i)].psw << endl;
		}
	}
}