PAT A1035 Password (20 分) 字符串_pat a1035 gets-优快云博客

本文链接：https://blog.youkuaiyun.com/chch1996/article/details/100087377

本文介绍了一种字符串处理算法，用于判断并替换指定字符。通过使用C++编程语言，实现了一个用户账号管理系统，该系统能够识别并修改包含特定字符的密码，确保密码安全性。

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题目大意：判断给出的字符串中是否有指定的字符，如果有，替换掉这些字符并输出。

直接替换就好。

AC代码：

#include <iostream>
#include <vector>
#include <map>

using namespace std;

bool modified(string &str, map<char, char>&mp)
{
    bool ret = false;
    for (int i = 0; i < str.size(); ++i)
    {
        if(str[i] == '1' || str[i] == '0' || str[i] == 'l' || str[i] == 'O')
        {
            str[i] = mp[str[i]];
            ret = true;
        }
    }
    return ret;
}

struct user
{
    string id;
    string password;
    user(string id, string password):id(id), password(password){};
};

int main()
{
    map<char, char>mp;
    mp['1'] = '@';
    mp['0'] = '%';
    mp['l'] = 'L';
    mp['O'] = 'o';
    int N;
    cin >> N;
    vector<user> v;
    for (int i = 0; i < N; ++i)
    {
        string id, password;
        cin >> id >> password;
        if(modified(password,mp)) v.push_back(user(id, password));
    }
    if(v.size() > 0)
    {
        cout << v.size() << endl;
        for (int i = 0; i < v.size(); ++i)
        {
            cout << v[i].id << " " << v[i].password << endl;
        }
    }
    else
    {
        if(N == 1) cout << "There is 1 account and no account is modified";
        else cout << "There are " << N << " accounts and no account is modified";
    }
    return 0;
}