Players can go all in before the cards are dealt, when they bet ante and blinds. Example of such a test:
100 200 50
5 1 1
preflop
1 1000
2 125
3 225
4 1000
5 25
0
950 0 0 750 0

Answer:
1
4 calls 200

a, b, c = map(int, input().split(" "))
a = int(a)
b = int(b)
c = int(c)
lst = [a, b, c]
lst.sort()
d = lst[0]-lst[1]*lst[2]
e = lst[0]-lst[1]-lst[2]
if d<e:
    min = d
else:
    min = e
print(min)

this test may help you:
4
0 0
1 0
2 0
3 0
2
1 2 1 0
1 0 1 0 1 19 1 0

ans:
2 0

"If there are several such propositions, output the one with the maximal number."

1500
1 1 -1000000000
1 1 -1000000000
1 1 -1000000000
1 1 -1000000000
1 1 -1000000000
1 1 -1000000000
1 1 -1000000000
1 1 -1000000000
.....

ans:
1 1
1 1

4
ab
ba
aba
bab

answer: ababa

Here is a simple "bad" tests generator on C++ to test your solutions (through comparing answers with brutforce solution):

const double PI = 3.14159265358979323846;
mt19937 rnd;

struct Point {
    double x, y;
};

double dist(const Point& a, const Point& b) {
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

bool check(const vector<Point>& points, const Point& p) {
    for (const auto& p2 : points)
        if (dist(p, p2) < 1e-3)
            return 0;
    return 1;
}

pair<vector<Point>, vector<Point>> gen_bad_test(int m = 100000, int n = 10000, double R = 10000, double r = 10) {
    vector<Point> cockroaches, sweets;
    cockroaches.reserve(m);
    sweets.reserve(n);
    while (m--) {
        Point p;
        do {
            double a = uniform_real_distribution<double>(0, 2*PI)(rnd);
            p = {cos(a) * R, sin(a) * R};
        } while (!check(cockroaches, p));
        cockroaches.emplace_back(p);
    }
    while (n--) {
        Point p;
        do {
            double a = uniform_real_distribution<double>(0, 2*PI)(rnd);
            double d = sqrt(uniform_real_distribution<double>(0, 1)(rnd)) * r;
            p = {cos(a) * d, sin(a) * d};
        } while (!check(cockroaches, p));
        sweets.emplace_back(p);
    }
    return {cockroaches, sweets};
}

This generator uniformly distributes points-cockroaches on a circle of radius R and points-sweets inside a circle of radius r.

And another generator that just distributes all points within [-MAX; MAX] along both axes:

pair<vector<Point>, vector<Point>> gen_rand_test(int m = 100000, int n = 10000, double MAX = 10000) {
    vector<Point> cockroaches, sweets;
    cockroaches.reserve(m);
    sweets.reserve(n);
    auto urd = uniform_real_distribution<double>(-MAX, MAX);
    while (m--) {
        Point p;
        do p.x = urd(rnd), p.y = urd(rnd);
        while (!check(cockroaches, p));
        cockroaches.emplace_back(p);
    }
    while (n--) {
        Point p;
        do p.x = urd(rnd), p.y = urd(rnd);
        while (!check(sweets, p));
        sweets.emplace_back(p);
    }
    return {cockroaches, sweets};
}

I just got AC for this problem and these two generators greatly helped me debug the solution revealing bugs and precision issues

1 0 2
-1 0 2

answer: 0

Не вздумайте решать задачу, оно того не стоит.

1
s

answer: 4

Don't forget to add primes <= k

how can it be improved????

n = int(input())
sp = []
a = []
c = 0
for i in range(n):
    sp.append(int(input()))
m = int(input())
for i in range(m):
    x = int(input())
    if x in sp:
        c += 1
print(c)

if a = 0 and b = 0 => r = 0 and q every complex number => count solation equation = infinity

realy

I would like to know if I can reset my progress here. I want to start afresh, but as I don't see a way to delete accounts, I thought that, at least, resetting my progress would be enough, but I can't do that either!

Thanks in advance for your help!

Edited by author 14.02.2025 08:40

int main(){
    int x, y;
    cin >> x >> y;
    if ((x*y-1)%2==1)
        cout << "[:=[first]";
    else cout << "[second]=:]";
    return 0;
}

set epsilon = 1e-9