Read article: convex hull of n*(n-1)/2 intersection points of n lines

Implementation is very easy.
But the proof is cool to read.

spell the months correctly

try kookko
ans: kook

#include <iostream>
#include <limits>
#include <cmath>

bool isPrime(int n) {
    if (n <= 1) {
        return false;
    }

    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            return false;
        }
    }

    return true;
}

int lowest_trivia_number (int I, int J) {
    double min_trivia = std::numeric_limits<double>::max();
    int res = 0;
    for (int i = I; i <= J; i++) {
        int sum_of_own_denominators = 1;
        for(int j = 2; j <= std::sqrt(i); j++){
            if(i % j == 0) {
                sum_of_own_denominators += j;
                if(j != i / j) {
                    sum_of_own_denominators += i / j;
                }
            }
        }

        double cur_trivia = sum_of_own_denominators/i;
        if (cur_trivia < min_trivia) {
            min_trivia = cur_trivia;
            res = i;
        }
    }
    return res;
}

int main() {
    int I, J;

    std::cin >> I >> J;
    if(I == 1) {
        std::cout << 1;
        return 0;
    }

    int largestPrime = -1;

    for (int num = J; num >= I; num--) {
        if (isPrime(num)) {
            largestPrime = num;
            break;
        }
    }

    if (largestPrime != -1) {
        std::cout << largestPrime << std::endl;
    } else {
        std::cout << lowest_trivia_number(I, J);
    }

    return 0;
}

Although can be solved in O(N^3)
Here's the interesting paper: Optimal Area Triangulation
https://core.ac.uk/download/pdf/226134402.pdf
(haven't read yet)

Hint: Also include O(N^3) solution

I precalculated digits for segment from 1e5 to 2e5, then i computed all digits from 1 to 1e5(if n >= 1e5), then i stepped from 1e5 to n with step size = 1e5. And after all this I computed digits from last step to n. It does about 1e9/1e5 steps and works 0.015 sec.

I think that it's dumb solution, so I want to know, how to solve this task in O(log10(n)) or smth similar.

I wrote two programs, one AC, one WA.
I run these two programs for many times but they always make
the same answer.
Is it strange?

a= input().split()

l = int(a[0])
h = int(a[1])
ω = int(a[2])

g = 9.81
ω = ω * 2 * 3.14159 / 60
t = (h / 100) / (g * 0.5)
x = ω * t * l

if x % l <= l/2:
    print("butter")
else:
    print("bread")

why do I get runtime error?
n = int(input())
for i in range (n) :
    a=list (map (int,input().split()))
a.sort()
c = a[::-1]
d = c[0] + c[1] + c[2]
e = (len(c) // 2)
print(d, e)

Time limit. Where is my mistake?
cnt1 = int(input())
lst1 = input()
lst1 = lst1.split()
cnt2 = int(input())
lst2 = input()
lst2 = lst2.split()
cnt3 = int(input())
lst3 = input()
lst3 = lst3.split()
cnt = 0
for i in range(cnt1):
    for j in range(cnt2):
        if lst1[i] == lst2[j]:
             for k in range(cnt3):
                    if lst1[i] == lst3[k]:
                        cnt += 1
        else:
             continue
print(cnt)

Edited by author 31.10.2017 14:42

Vectors calculations uses square roots. Could this lead to not very precise floats and wrong results?

Edited by author 05.10.2023 13:50

Edited by author 05.10.2023 13:50

I just don't know what is wrong.I need some test in order to pass test 8

Check  that you indeed turn after reaching point B, i.e. that intersection point is not between A and B or you don't even reach B.

it is said that the testcase 9 is as follow：
4 4 1 11.1
1 2 2.0 0.5 0.5 0.5
2 3 1.05 1 0.01 1.0
3 4 1.05 1 0.01 1.0
4 2 1.05 1 0.01 1.0

the right answer is YES.

but i think there is no possibility to increase his money, so the answer should be NO.
could someone explain this test case to me.? thanks a lot..

Edited by author 19.09.2010 18:45

At the beginning of the turn, does player choose only among the parallelepipeds generated during the last turn, or among all of them, including the previous turns?

For example, K=1, n1=11

First player dissects it at x=2. We have now two parallelepipeds, of size 1 and 9.
Second player chooses the one of size 9, and dissects it at x=3, generating parallelepipeds of size 2 and 6.
Does first player have choice between 1, 2 and 6? Or only 2 and 6?

Thanks to authors :)

Try this test:
10 100 50 10 10 10 99
1 1 1 1 1 1 1 1 1 1

#pragma GCC optimize("Ofast")
With this line of code I'm getting AC 0.468, without it I'm getting TL14

give me test,pleasee