Is it possible to use somehow that p is prime?
My AC solution doesn't use this fact, but I'm wondering is there a simpler way to solve this task.

Hi I think that we should help each other. If someone resolves a problem he/she must help the others with materials or links where the algorithm/formula is explained. This is the reason we are users of such sites - because we want to improve our programming skills.
So I decided to help all of you who find this problem difficult. Here you can find help:

http://en.wikipedia.org/wiki/Circular_segment

http://www.mathopenref.com/segmentarea.html

The rest is up to you!
HTH

i checked it manually and answer was correct. Why does if fail?
Code:
#include <iostream>


int main(){
    std::string lock1, lock2;
    std::string result = "yes";
    int code = 1;
    int lock_val1, lock_val2, cur_value;
    bool lock_state = false;
    std::cin>>lock1;
    std::cin>>lock2;
    lock_val1 = std::stoi(lock1);
    lock_val2 = std::stoi(lock2);
    while (code < 10000)
    {
        if (lock_state == true){
            cur_value = lock_val1;
        }
        else{
            cur_value = lock_val2;
        }

        if (code > lock_val1 && code > lock_val2){
            result = "no";
            break;
        }

        if (code == cur_value){
            break;
        }

        lock_state = !lock_state;
        ++code;
    }

    std::cout << result << std::endl;
    return 0;
}

Edited by author 30.11.2021 13:19

Edited by author 30.11.2021 13:19

Edited by author 30.11.2021 13:24

Edited by author 30.11.2021 13:37

When i used dfs, i was getting "Division by zero" on 10 test. Then' i switched to bfs and got AC. Can you explain me: WHY?

The grid is m*n not n*m

I have wa in 7 test

Length of the first string can be bigger than length of the second string. I spent 15 submissions to realise this

1) First test is not sample
2) There are not n strings in input

So, if you have a number, that divides 2^n, than you can add some numbers in the begining (not more than 10) to get number, that divides 2^(n+1). That means, you can bruteforce all possible numbers, that you can add in the begining. Works in O(n*2^10)

I've tried to submit the same code with different compilers and have received "compilation error" from MS Visual C++ compilers ans "accepted" from G++ and Clang++.

On my laptop with MS VS 2019 this code also runs.

The question is are compilers on the server ok?

the store had the product, but sold it = the store did not have the product
if (!Products.count(name) || !Products[name]) Purchases.erase(Purchases.begin() + i);

Test:
1
10 of gg
3
1 of ff
10 of gg
9 of gg

Correct Answer: 3

Use python

You can always fill 2 * m area

My program got WA1 with answer
1 2 3 4
1 2 3 4
on example test
Why?

tle 42 -ios_base::sync_with_stdio(0);
but ios_base::sync_with_stdio(false); cin.tie(NULL); - ACCEPTED with 0.015)

Is there a solution without randge tree structure?

dp[i] - минимальное количество отрезков, если последний отрезок заканчивался в позиции i. ответ dp[m].