题目描述

blablabla

样例

#include<iostream>
using namespace std;

const int N=1010;
string name[N],pwd[N];
int m=0;

string change(string s)
{
    string str;
    for(auto c:s)
    {
        if(c == '1')    str+='@';
        else if(c == '0')    str+='%';
        else if(c == 'l')    str+='L';
        else if(c == 'O')    str+='o';
        else    str+=c;
    }
    return str;
}

int main()
{
    int n;
    cin>>n;
    string cur_id,cur_pwd;
    string change_pwd;
    for(int i=0;i < n;i++)
    {
        cin>>cur_id>>cur_pwd;
        change_pwd=change(cur_pwd);
        //cout<<change_pwd<<" "<<"111";    
        if(change_pwd != cur_pwd)
        {
            name[m]=cur_id;
            pwd[m]=change_pwd;
            m++;
        }
    }

    if(m == 0)
    {
        if(n != 1)  printf("There are %d accounts and no account is modified\n",n);
        else printf("There is 1 account and no account is modified");
    }
    else
    {
        cout<<m<<endl;
        for(int i=0;i < m;i++)
        {
            cout<<name[i]<<" "<<pwd[i]<<endl;
        }
    }
    return 0;
}

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla