Hi!

I believe test #3 is invalid - there is no way to get from floor 1 to floor N.

Can you please check?

Thank you!

Tricky test case (even can't find its analogue). It contains such placement of draughts that you should precisely touch one or several other draughts by your move, and almost identical angles (+/- 1e-8 radians) will not give correct answer. When allow precisely touch got AC.

P.S.
Seems like one enemy draught you touch on the left side and other one - on the right side simultaneously.

Edited by author 23.04.2023 18:44

Is this a rounding issue in the output?

input:
804289384 2
36 94
795951522 804289383

output:
8556270

631 8
902 694
854 198
479 378
808 647
53 134
77 91
493 13
597 802
ans: 22

86 15
27 93
95 30
85 88
26 17
78 66
49 60
52 21
96 85
55 45
2 69
1 47
18 20
96 80
66 8
70 51
ans: 22

75 3
49 97
16 10
10 36
ans: 8

be more careful about formulas

If the middle nail was hammered, then the left and right become adjacent?
Answer: NO

Edited by author 10.03.2026 13:47

just overrated

The vectors may have negative coordinates.

|x|, |y| > 2^32

Я написал рекурсию которая каждый раз делила отрезок на два и брала максимальный среди ответа всех таких отрезков которых поделила.(типо Merge Sort). У меня был memory limit на 3 тесте. Это значить рекурсия берет память?

n,k = map(int,input().split())
if k==1 or n==1:
    print(2*n)
else:
    if 2*n%k==0:
        print(2*n//k)
    else:
        print(2*n//k+1)

N,M  = map(int,input().split())
k = []
for _ in range(M):
    m = int(input())
    k.append(m)
for i in range(1,N+1):
    print(f"{k.count(i)*100/M:.2f}%")
Time limit please help

What does it mean: "reach smth"?
It's a strict condition, but how should i understand it?)
In my opinion, this sample has two right answers:
4
0 0
0 1
0 5
0 8
2 * (1.5 * 1.5) * M_PI or (1 + 3 * 3) * M_PI.

input:
2 8
12 28 40 46
48 54 60 80
2 7
9 24
34 49
54 62
66 77
78 79
80 83
84 84

output:
6

input #1:
3
110

output #1:
YES

input #2:
4
1110

output #2:
NO

Edited by author 24.02.2026 17:26

12
12 5 3 11 2 8 4 9 1 6 10 7

answer: 34650  in example output.

But, following full check test prints 6300.

-----
static bool match(int a[], int n, int b[], int m){
    if (n != m) return false;
    if (n == 0 && m == 0) return true;

    int i = std::max_element(a, a + n) - a;
    int j = std::max_element(b, b + m) - b;

    return i == j && match(a, i, b, j) && match(a + i + 1, n - i - 1, b + j + 1, m - j - 1);
}
int main()
{
        int n = 12;
        int a[] = { 12, 5, 3, 11, 2, 8, 4, 9, 1, 6, 10, 7 };
        int b[] = {  1, 2, 3,  4, 5, 6, 7, 8, 9, 10, 11, 12};
        int ans = 0;
        do{
            if (match(a, n, b, n)) ++ ans;

        } while (std::next_permutation(b, b + 12) );

        printf("ans = %d\n", ans);
        return 0;
}

I solved it using pretty straigtforward DP O(n * m^2), obviously it's just a complexity without any additional factors but it doesn't seem to be fitting time limit. Somehow it did. But I wonder, is there any more intricate idea for a faster solution?

try this test:
1 1 3 100 3000

ans: 1000

how to know when there are no more inputs