I bruteforced for all integers in range(-2000,2000) and got the best approximation for value a1. Then I brutforced for all doubles in range(a1-100;a1+100) and got AC

Try this:
1 5
1.00
answer: 0.20

try longdouble XD
I used longdouble instead of double and got AC

My sulution a has n^2 * 2 ^ n and get TL
accepted solution supposed to be O(n * 2^n), right?

I have some tests for my program:
N    K    Result
3    2    4
4    3    5
4    2    8
5    2    12
5    3    8
5    4    6
6    2    18
6    3    12
6    4    8
6    5    7
7    2    24
7    3    16
7    4    12
7    5    9
7    6    8
8    2    32
8    3    21
8    4    16
8    5    12
Check my answers, please!
                        Thank you.

Take care in the cases of multiple lines. I was thinking it wasn't possible and it was giving me WA5. When I took care of it I got AC.

Why I have WA?

#include <iostream>
#include <cmath>

using namespace std;

int main() {
    double const pi = 3.14159265358979323846;
    double a;
    double r;
    cin >> a >> r;
    long double s = pi * r * r;
    if (r <= a / 2) {
        cout << roundl(s * 1000) / 1000;
    } else if (r >= ((a / 2) * sqrt(2))) {
        cout << (int) (a * a) << endl;
    } else {
        double cosx;
        cosx = (a / 2) / r;
        double rad;
        rad = acos(cosx) * 2;
        double segm = (r * r / 2) * (rad - sin(rad));
        s = s - 4 * segm;
        cout << round(s * 1000) / 1000 << endl;
    }
    return 0;
}

Solved with Z-function in 17 rows of code (if don't count 10 rows for Z-function)

I've got AC
My heap was so small...(Could anyone talls me why 400 is unenough?)

Edited by author 22.09.2021 17:55

If you use mask as two-dimensional array you probably will get TLE, instead of this use bitset from STL

Edited by author 20.09.2021 15:25

5
2 4 3 1 5

Not a proof

5
2 4 1 3 5

Cheater

1 0 1 0 0 0
11
ans = 10

n/=2;
dp[n][n*9];
dp[i][j] = dp[i-1][j]+dp[i-1][j-1] + ... + dp[i-1][j-9];
ans = dp[n][0]*dp[n][0] + ... + dp[n][n*9]*dp[n][n*9];

Edited by author 17.09.2021 19:12

Edited by author 17.09.2021 19:12

Precalculate pythagorean tripples

9
3 9 8 2
3 7 3 1
2 4 2
2 5 3
3 7 6 4
3 9 8 5
2 5 2
2 6 1
2 6 1
Please try this data,please
My progame can't pass this,but I got AC

My code:
n,m = map(int, input().split())
dct = dict()
for i in range(m):
    a = input()
    if a not in dct:
        dct[a] = 1
    else:
        dct[a] += 1
for i in dct:
    print("{0:.2f}%".format((dct[i]/m)*100))


Where is mistake?

Edited by author 13.09.2021 21:32

Hint: you should only know fundamental theorem of arithmetic

Треугольное число

The question stated that 0 < N < 30.  However, in test case #9 N is exactly 30, contradicting the constraint.