Becuase of edge cases,this problem take me many hours.
So I put one correct sorting function here.
struct ND{// Single bracket sequence
    int len,mn=0;
    int cnt=0;
    int id;
// len is the length of the bracket sequence
// if we define '(' as +1,and ')' as -1,the minimum prefix sum of the bracket sequence is mn.
// and the total sum of the sequence is cnt
// id is the index of the bracket sequence in the given input.
}nd[maxn];
bool swap(ND& a,ND& b){
        if(a.cnt>=0&&b.cnt<0)return 1;// a first
        if(a.cnt<0&&b.cnt>=0)return 0;// b first
        if(a.cnt>=0)return -a.mn<-b.mn;
        return -b.mn+b.cnt<-a.mn+a.cnt;
}

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

Any solution with time complexity strictly lower than O(nmk) will be helpful. Thanks!

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

Is this a rounding issue in the output?

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