#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

It's kinda tedious since you need most operations too

tell me the hint, i want to solve this problem.