What is in test7?


Edited by author 23.02.2023 13:59

l1=[]
l2=[]
l3=[]
for i in range(0,1000):
    l3.append(False)
def funct(a):
    b=a.split()
    if b[0]=='register':
        if b[1] not in l1:
            l1.append(b[1])
            l2.append(b[2])
            print("success: new user added")
        else:
            print("fail: user already exists")
    elif b[0]=="login":
        if b[1] not in l1:
            print("fail: no such user")
        else:
            k=l1.index(b[1])
            if l3[k]:
                print("fail: already logged in")
            else:
                if l2[k]==b[2]:
                    l3[k]=True
                    print("success: user logged in")
                else:
                    print("fail: incorrect password")
    elif b[0]=="logout":
        k=l1.index(b[1])
        if l3[k]:
            l3[k]=False
            print("success: user logged out")
        else:
            print("fail: already logged out")
f=int(input())
for m in range(0,f):
    n=input()
    funct(n)





f**king time limit!!!

in some cases, there are more than 1 current user

test 4 ??? help me please

test 3 ??? help me please

Please help with wrong answer 7.
Possible input?

please can anyone tell me what is the test of 5.

I tried to solve it with C++.  It was long.  It was PAINFUL!  Finally,  I didn't managed to solve it in C++.  And then I just opened my Python IDE and solves it within 10 minutes.  Just AC first attempt.  The task is evil.

ll n, k, op, a[11][11], ans;
string s;
vector < pair < string, string > > alr;
vector < pair < string, string > > lg;
vector < string > out;
int main(){
    cin >> n;
    for(ll i = 1; i <= n; i++){
        cin >> s;
        k = 0;
        if(s == "register"){
            string lo, pa;
            cin >> lo >> pa;
            for(ll i = 0; i < lg.size(); i++){
                if(lo == lg[i].first){
                    cout << "fail: user already exists" << "\n";
                    k = 3;
                    break;
                }
            }
            if(k != 3){
                lg.pb(mp(lo, pa));
                cout << "success: new user added" << "\n";
            }
        }
        k = 0;
        if(s == "login"){
            string acc, par;
            cin >> acc >> par;
            for(ll i = 0; i < alr.size(); i++){
                if(acc == alr[i].first){
                    if(par == alr[i].second){
                        cout << "fail: already logged in";
                        k = 2;
                        break;
                    }
                }
            }
            for(ll i = 0; i < lg.size(); i++){
                if(acc == lg[i].first && k != 2){
                    ans++;
                    if(par == lg[i].second){
                        cout << "success: user logged in";
                        alr.pb(mp(acc, par));
                        out.pb(acc);
                        k = 1;
                        break;
                    }
                }
            }
            if(!k && ans == 0)
                cout << "fail: no such user";
            if(!k && ans > 0)
                cout << "fail: incorrect password";
            cout << "\n";
        }
        ans = 0;
        k = 0;
        if(s == "logout"){
            string avt;
            ll o;
            cin >> avt;
            for(ll i = 0; i < out.size(); i++){
                if(avt == out[i]){
                    out[i] = "````````21";
                    k = 4;
                    break;
                }
            }
            op = 0;
            for(ll i = 0; i < lg.size(); i++){
                if(avt == lg[i].first){
                    op = 1;
                    break;
                }

            }
            if(k != 4 && op){
                cout << "fail: already logged out";
            }
            else
                if(op)
                cout << "success: user logged out";
            if(!op)
                cout << "fail: no such user";
            cout << "\n";
        }
        op = 0;
        k = 0;
    }
}





what is wrong???

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Timus2002 {

    public static void main(String[] args) {

        Map<String, String> map = new HashMap<>();
        Scanner sc = new Scanner(System.in);
        String[] s = new String[105];
        ArrayList<String> logged = new ArrayList<>();
        int n = Integer.parseInt(sc.nextLine());
        for (int i = 0; i < n; i++) {
            s[i] = sc.nextLine();
        }

        for (int i = 0; i < n; i++) {
            String[] mas = s[i].split(" ");
            if(mas[0].equals("register")){
                if(map.containsKey(mas[1])){
                    System.out.println("fail: user already exists");
                    continue;
                }
                map.put(mas[1], mas[2]);
                System.out.println("success: new user added");
            }

            String login = mas[1];
            if(mas[0].equals("login")){
                String pass = mas[2];
                if(!map.containsKey(login)) {
                    System.out.println("fail: no such user");
                    continue;
                }
                String value = map.get(login);
                if(!value.equals(pass))
                    System.out.println("fail: incorrect password");

                if(value.equals(pass)) {
                    if (!logged.contains(login)) {
                        logged.add(login);
                        System.out.println("success: user logged in");
                    }
                    else
                        System.out.println("fail: already logged in");

                }
            }

            if(mas[0].equals("logout")) {
                if (!map.containsKey(login))
                    System.out.println("fail: no such user");
                if (!logged.contains(login))
                    System.out.println("fail: already logged out");
                else {
                    System.out.println("success: user logged out");
                    logged.remove(login);
                }
            }
        }
    }
}

Help me please.. Why WA 3?
I "Use this letter as a formal description of the algorithm and follow the described format of system messages" and still get Wrong Answer..

This test is correct?

7
login a 1
fail: no such user
logout a
fail: no such user
register a 1
success: new user added
login a 1
success: user logged in
login a 1
fail: already logged in
login a 2
fail: incorrect password
register a 2
fail: user already exist

Maybe you tell a counterexample or hint?

Thanks.

Edited by author 02.08.2016 21:56

Can anyone tell me  the "main idea" of test 4 or type test. I got AV. Need some reason where is my mistake (to my mind - arrays, but need help). Thanks.

switch(in.sval)
            {
                case "register":
                    {
                        in.nextToken(); log=in.sval; if(log==null) log=Integer.toString((int)in.nval);
                        in.nextToken(); pas=in.sval; if(pas==null) pas=Integer.toString((int)in.nval);
                        if(m.containsKey(log)) out.println("fail: user already exists");
                        else
                        {
                            m.put(log,pas);
                            out.println("success: new user added");
                        }
                    }
                    break;
                case "login":
                    {
                        in.nextToken(); log=in.sval; if(log==null) log=Integer.toString((int)in.nval);
                        in.nextToken(); pas=in.sval; if(pas==null) pas=Integer.toString((int)in.nval);
                        if(!m.containsKey(log)) out.println("fail: no such user");
                        else if(!m.get(log).equals(pas)) out.println("fail: incorrect password");
                        else if(a.contains(log)) out.println("fail: already logged in");
                        else
                        {
                            a.add(log); //out.println(m.get(log) + " " + pas);
                            out.println("success: user logged in");
                        }
                    }
                    break;
                case "logout":
                    {
                        in.nextToken(); log=in.sval; if(log==null) log=Integer.toString((int)in.nval);
                        if(!m.containsKey(log)) out.println("fail: no such user");
                        else if(!a.contains(log)) out.println("fail: already logged out");
                        else
                        {
                            a.remove(log);
                            out.println("success: user logged out");
                        }
                    }
                    break;

            }

please can anyone tell me what is the test of 4?????

#include <iostream>

#include <conio.h>

using namespace std;

struct abonent
{
    char login[30];
    char password[30];
    short status;
};
int main()
{
    short n, k = 0, j, compare, index;
    char *part_str;
    char str[80];
    cin >> n;
    abonent *peop = new abonent[n];
    gets(str);
    for (int i = 0; i < n; i++)
    {
        gets(str);
        part_str = strtok(str, " ");
        if (!(strcmp(part_str, "register")))
        {
            part_str = strtok(NULL, " ");
            strcpy(peop[k].login, part_str);
            compare = 1;
            for (j = 0; j < k; j++)
            {
                if (!(strcmp(peop[k].login, peop[j].login)))
                {
                    compare = 0;
                    break;
                }
            }
            if (compare)
            {
                part_str = strtok(NULL, " ");
                strcpy(peop[k].password, part_str);
                peop[k].status = 0;
                k++;
                cout << "success: new user added\n";
            }
            else
            {
                cout << "fail: user already exists\n";
            }
        }

        if (!(strcmp(part_str, "login")))
        {
            part_str = strtok(NULL, " ");
            compare = 1;
            for (j = 0; j < k; j++)
            {
                if (!(strcmp(part_str, peop[j].login)))
                {
                    compare = 0;
                    index = j;
                    break;
                }
            }
            if (!compare)
            {
                part_str = strtok(NULL, " ");
                if (!(strcmp(part_str, peop[index].password)))
                {
                    if (peop[index].status)
                    {
                        cout << "fail: already logged in\n";
                    }
                    else
                    {
                        peop[index].status = 1;
                        cout << "success: user logged in\n";
                    }
                }
                else
                {
                    cout << "fail: incorrect password\n";
                }
            }
            else
            {
                cout << "fail: no such user\n";
            }
        }

        if (!(strcmp(part_str, "logout")))
        {
            part_str = strtok(NULL, " ");
            compare = 1;
            for (j = 0; j < k; j++)
            {
                if (!(strcmp(part_str, peop[j].login)))
                {
                    compare = 0;
                    index = j;
                    break;
                }
            }
            if (!compare)
            {
                if (peop[index].status)
                {
                    peop[index].status = 0;
                    cout << "success: user logged out\n";
                }
                else
                {
                    cout << "fail: already logged out\n";
                }
            }
            else
            {
                cout << "fail: no such user\n";
            }
        }
    }
    delete[] peop;
    return 0;
}

Edited by author 24.02.2015 02:52

I don't understand why WA #1 because I fully tested answer of mu programm and answer on site. They hasn't got differences! Even invisible characters. Here is my code:

var n,i,j,fl,k,n1:longint; a,b,d,x:string; y,z:array [1..100] of string;
z1:array [1..100] of byte; c:char;
begin
readln (n);
for i:=1 to n do begin
fl:=0;
a:=''; b:=''; d:='';
readln (x);
n1:=length (x);
j:=0;
repeat
j:=j+1;
If x[j]<>' ' then a:=a+x[j];
until x[j]=' ';
repeat
j:=j+1;
If x[j]<>' ' then b:=b+x[j];
until x[j]=' ';
repeat
j:=j+1;
If x[j]<>' ' then d:=d+x[j];
until j>=n1;
fl:=0;
If a='register' then begin
for j:=1 to k do
If b=y[j] then fl:=1;
If fl=1 then writeln ('fail: user already exists') else begin
writeln ('success: new user added'); k:=k+1; y[k]:=b; z[k]:=d; z1[k]:=0;end;end;
fl:=0;
If a='login' then begin
for j:=1 to k do
If b=y[j] then fl:=j;
If fl=0 then writeln ('fail: no such user') else
If (fl>0) and (d<>z[fl]) then writeln ('fail: incorrect password') else
If (fl>0) and (d=z[fl]) and (z1[fl]=1) then writeln ('fail: already logged in') else begin
writeln ('success: user logged in'); z1[fl]:=1;end;end;
fl:=0;
If a='logout' then begin
for j:=1 to k do
If b=y[j] then fl:=j;
If fl=0 then writeln ('fail: no such user') else
If (fl>0) and (z1[fl]=0) then writeln ('fail: already logged out') else begin
writeln ('success: user logged out'); z1[fl]:=0;end;end;end;
end.

more than one registrations ;)

Сan somebody tell whether it may be several current users at once?
Может кто нибудь сказать, может ли быть несколько текущих пользователей сразу?

Edited by author 01.02.2014 02:35

Is n no more than 100?
The time complexity of my program is O(n*n), why does it take so much time to run my program?(0.531s)
Do we need O(n*log(n)) to solve this problem?