AC QH 10H
JC KC 4D
AC AC

4D 5D AD
4C 5C JC
2H AC

5C 5D AD
5S 5H JC
2H AC

5C 5D AD
5S 5H JC
2H AD

5C 5D AD
5S 5H JC
JH JD

JS 8H JC
8C 8D 10H
8H 10H

10D 8D 9D
3D 3C 3H
AH AS

2C 3D 4H
2D 3H 4S
AH AH

8C QC JC
JD 9D KD
3H 3H

8C QC KC
JD 9D KD
3H 3H

10C QC KC
JD 9D KD
3H 3H

5C 10C 10D
10S 9D 6D
3H 3H

AC 10C 10D
10S 9D 6D
3H 3H

AC 10C 9C
10S 9D 6D
3H 3H

AC 10C 9C
10S 10D 6D
3H 4S

AH AS AD
AC 10C 10S
3H 4S

AH AS AD
AC 10C JC
3H 4S

AH AS AD
AC 10C JC
3H 4S

AH QS KD
AD KC JD
3H 4S

AH QS KD
AD KC JD
3H 4S

AH AS KD
AD AC QD
3H 4S

AH AS 3D
KS KC 2D
3H 4S

AH AS 3D
KS KC 2D
2H 2S

AH AS 3D
KS KC KD
2H 3H

AH AS AD
KS KC KD
2H 3H

3S 4C 5C
2S 4S 5S
AH AD

3S 4C 5C
AS AH 2C
AH AD

3C 4C 5C
AS AH 2C
AH AD

3C 4C 5C
2S 7S 10S
AH AH

4S 5D 8H
5S 7S 8D
3C 2D

4S 4D 4H
5S 5D 2H
2C 2D

4S 4D 4H
5S 5D 3H
3C 3D

4S 4D 4H
5S 5D 6H
6C 6D

4S 4D 4H
5S 5D 5H
5H 5H

AH AS JS
AC AD 10S
KS KD

AH AS KH
AC AD 10S
KS KD

3C 4C 5C
4S 5S 6S
AH AH

3C 4C 5C
3S 4H 5D
AH AH

2S 4S 5S
7D 4D 5D
AH AH

3D 4D 5D
2S 3S 4S
AH AH

4C 4D JS
4S 4H 3H
7S 7S

4C AC 2C
4S KD 6H
5C KH

JS 8H JC
8C 8D 10H
8H 10H

10D 8D 9D
3D 3C 3H
AH AS

2C 3D 4H
2D 3H 4S
AH AH

Sasha
Dima
Artyom
Sasha
Dima
Sasha
Dima
Artyom
Dima
Sasha
Sasha
Sasha
Sasha
Sasha
Sasha
Sasha
Sasha
Sasha
Sasha
Sasha
Sasha
Dima
Dima
Dima
Sasha
Sasha
Dima
Sasha
Sasha
Dima
Sasha
Dima
Dima
Dima
Dima
Artyom
Dima
Sasha
Dima
Sasha
Sasha
Dima
Sasha
Dima
Artyom

The output needs to be sorted in ascending order(!)

so for test number 5: 45

the correct answer is 59 (5*9)

and 95 (9*5) will be rejected.

Hope this helps someone.

1 12
1 1 10 8
1 1 8 10
1 1 9 3
1 1 7 4
1 1 2 4
1 1 5 4
1 1 2 5
1 1 9 9
1 1 3 5
1 1 9 2
1 1 9 5
1 1 5 9
1 1 6 4

Ans:
1
4

Also, check that you print "Impossilbe", when it's impossible to achieve the finish vertex

I have correctly performed all tests from the forum, but still WA 3. Please tell me what the problem is or write new tests for me.

solution on the surface, take a look at the arithmetic mean P

#include<bits/stdc++.h>
using namespace std;
int main(){
    int t;
    cin >> t;
    while(t--){
        string s;
        cin >> s;
        int a = int(s[1] - 48);
        if(s[0]=='a' ||s[0]=='h'){
            if(a==1 ||a==8)
                cout << 2 << endl;
            else if(a==2||a==7)
                cout << 3 << endl;
            else if(a>2)
                cout << 4 << endl;
        }
        else if(s[0]=='b'||s[0]=='g'){
            if (a == 1 || a==8)
                cout << 3 << endl;
            else if (a == 2|| a==7)
                cout << 4 << endl;
            else if (a > 2)
                cout << 6 << endl;
        }
        else{
            if (a == 1|| a==8)
                cout << 4 << endl;
            else if (a == 2|| a==7)
                cout << 6<< endl;
            else if (a > 2)
                cout << 8<< endl;
        }
    }
}

Из точки (2,2) мы можем попасть в (1,1),а потом в (0,0), сразу в (0,0), в (1,0),а потом в (0,0). В первом случае t=0.63246+0.23277=0.86522, во втором t=0.86522, в третьем t=0.70711+0.15811=0.86522.

Я использую формулу для t=(-v+sqrt(v*v+2*g*len*sina))/(g*sina)),если наклонный отрезок и t=len/v иначе.

see https://github.com/lma10h/big_int

I tested for a long time and tried to understand why the thymus mockingly shows me such a mocking number, it's all because of the portals, it's just impossible there, I got so confused in them that I didn't understand how my code worked, in general, how to avoid such a problem. add all sorts of checks near the portals, which is right wherever there are portals, just destroy these portals completely, absolutely

dont add negative numbers to answer

i think its something like this
2
ab\df\fd
df\df
ab
 df
  fd
df
 df
  fd

Сама задача конечно же довольно простая. Решив ее несколькими способами я каждый раз упирался в превышение объема допустимой памяти на 11 тесте. Соответственно львиная доля времени ушла на осознание того, что же именно потребляет память и на устранение этого узкого места.
Как можно было догадаться, память жрали Строки. Я убрал использование string отовсюду кроме считывания данных из консоли. Хранение данных реализовано через два массива int. Для разбора ввода и для вывода ответов в консоль использовались массивы int и char.

Вообще, это одна из тех задач, которые отлично иллюстрируют проблему, описанную в FAQ "Как писать решения на C#":
"В некоторых задачах потребуется собственная быстрая реализация разбора входных данных и форматирования выходных". Только здесь проблема не в скорости, а в памяти.

somebody faced this problem? somebody has a test data?

1

if time of djek ravno time of this people so, this is crash

i just draw its in paint and use some kinematick formulas and solve it by that

These two tests helped we find my bugs.

Test
2
1
3
0
2 1 2
The answer is 4.

Test
3
10
20
30
1
29 3 1 2 3
2 1 2
The answer is 29.

Hope it will help you too.

if you use the two-pointer method, then you have problems with the left pointer

N =int(input())
Nl = []

for i in range(N):
    Nl.append(int(input()))

M = int(input())
Ml = []

for j in range(M):
    Ml.append(int(input()))

answerl = []


Nl = set(Nl)
Nl = list(Nl)
Nl.sort()

for ni in Nl:
    for mi in Ml:
        if mi == ni:
            answerl.append(mi)
        else:
            continue

print(len(answerl))

#include <stdio.h>
    int main(){
        int a, b, c;
        scanf("%d", &a);
        b=12-a;
        if(b*45<=240){
            printf("Yes");
        }
        else{
            printf("No");
        }
        return 0;
    }