check the values ​​of y[i] by x[i] - x[i-1] in the straight line equation
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use ceil() and floor()

I'm confused by this sentence in the problem statement:
"the murderers have agreed that their testimonies will have no contradictions between them"

Does that mean that tests like this are invalid?
3
A 1 2
B 1 3
C 1 1

In this test there are 3 people and each one conflicts with at least one another. If one person is declared a murderer, there is still a contradiction between the other two.
No 2 people can be declared murderers because there is always a contradiction between any 2 of them.

Let's consider the case a = 0 and b = 0. Obviously, the answer will be n * n, because none of the knights will be able to move to any other cell of the table. Now let's assume that a != 0 or b != 0. By doing some casework, one can show that the answer for n will be the same as the answer for min(n, 10), so we can assume that n <= 10. Then, just find all possible night moves and get connected components of the corresponding graph. The answer will be the number of connected components.

if k > n my program worked with n < 0
you should to consider the case after cycle: if n < 0 then n = 0

Edited by author 03.04.2020 21:51

We cannot imagine the test where our prog will give wrong answer!

#include <bits/stdc++.h>

using namespace std;

int main(){
    int n;
    int a[100000];
    a[0] = 0;
    a[1] = 1;
    int max = 1;
    while(cin >> n, n != 0){
        if(n > max){
            for(int i = max + 1; i <= n; ++i){
                if(i % 2 == 0){
                    a[i] = a[i / 2];
                }
                else{
                    a[i] = a[(i - 1) / 2] + a[(i - 1) / 2 + 1];
                }
                max = n;
            }
        }
        cout << a[n - (n % 2 == 0)] << "\n";
    }
}

Edited by author 10.01.2021 03:55

24303989349327424 = 2^6 * 11^14 (Yes)
79792266297612001 = 7^20 (Yes)
467851794974552912 = 7^1 * 2^4 * 11^15 (Yes)
566608619280937216 = 2^8 * 19^12 (Yes)
932195579567617600 = 5^2 * 2^6 * 17^12 (Yes)
999983616067108864 = 1953109^2 * 2^18 (Yes)
999197664639844352 = 19681^3 * 2^17 (Yes)
998848442311376896 = 15619^3 * 2^18 (No)
999999999987679232 = 1907348632789^1 * 2^19 (Yes)
798352691495399040 = 2^7 * 3^1 * 5^1 * 7^1 * 11^1 * 13^1 * 17^1 * 19^1 * 23^1 * 29^1 * 31^1 * 37^1 * 41^2 (Yes)

a, b = int(input()), int(input())
print(a + b)

I've learned A LOT of optimizations tricks for BFS with this problem. I think that's the fastest BFS I've ever done LOL. A little hint if you are unsure what to do: Try to brute force the solution and optimize it. Eventually, you will get AC. And thanks to the author, for creating this amazing problem.

I use Disjoint-set
who can help me?

After so many calculation and math I have solved the problem.....
Here we can have two worst cases...

Case 1:
having all the right shoes first.so here needed time is 2*b and we have now all the left shoes remaining...so total time is 2*b+40...

Case 2:
we may have 39 right shoes so time needed here is (39*2=78)...then we have only one right foot left but we may encounter all the left shoes and here needed time is 40+2*(a-40)....> 40 for the first 40 shoes and 2*(a-40) is for the remaining shoes as they needed to be thrown away...then we have the only one right foot left and it need 1 second...
so total time = 78+40+2*(a-40)+1 = 119+2*a-80 = 2*a-39

ans=max(Case 1,Case 2)


Edited by author 22.02.2020 03:07

int f;
    cin >> f;
    int obs = 255;
    /*
    5 hours = 300 minutes
    all time = 300 minutes - 45 minutes(for f tasks)
    */
    int r = 12 - f;
    // tasks after first hour
    int ans = 45 * r;
    // time for solution
    if (ans <= obs) cout << "YES" << endl;
    else cout << "NO" << endl;

Edited by author 15.05.2020 17:20

Edited by author 15.05.2020 17:20

#+-+-+-+-+-+
+-+-+-+-+-+#

Edited by author 09.01.2021 01:54

My AC solution (6438921) does not work on this test

2
111111110000
6
1 9
2 3
3 4
5 11
6 7
7 8

Please add it. Moreover, I think there could be found much more tricky tests than the one above. Please try to do that and add them too (I am too lazy to help you with that :) ). You may look into my submission 6438944 where this case is handled (presumably as well as all other tricky cases).

I don't know the problem

use std::io;
use std::io::BufReader;
use std::io::BufRead;
fn minimal_product_digit(n: i64) -> i64{
    if n == 0 {return 10};
    if n < 10 { return n};

    let f : Vec<i64> = vec![2,3,5,7];
    let mut aux = n;
    let mut ret : Vec<i64> = Vec::new();
    for i in f{

        if  aux % i== 0 {

            loop{
                aux/=i;
                 ret.push(i);
                if aux % i != 0 {
                    break;
                }
            }
        }
    };

    if ret.is_empty() || aux > 1 {
        return -1;
    }

    match ret.iter().fold(0,|acc,elem| if *elem == 3{acc + 1} else {acc}) {

        1 => {

            let index = ret .iter().position(|&x| x == 3).unwrap();

            if index > 0{

                ret.remove(index);

                let k = ret.get_mut(index - 1).unwrap();

                *k = 6;

                ret.sort();

            }
        }

        _ => ()

    }
    let mut v : Vec<i64> = Vec::new();
    for i in ret{

        match i {

            2 | 3=>

                match v.last_mut(){

                    None => v.push(i),
                    Some(last) if i ==3 && *last == 2 => v.push(i),
                    Some(last) => {
                        let valor = *last;
                        if (valor * i) < 10{
                            *last = valor*i
                        }
                        else{
                            v.push(i);
                        }
                    }
                }
            _ => v.push(i)
        }
    }

        v.sort();
        v.iter().fold(0, |acc, elem| acc * 10 + elem)

}

fn main() {

    let mut line = String::new();
    let mut stdin = BufReader::new(io::stdin());

  stdin.read_line(&mut line).unwrap();

    let lim = line.trim().parse::<i64>().unwrap();

    println!("{:?}",minimal_product_digit(lim));

}

When n > 5 I guess, you can place one or more circles to free place within polygon?
I assume that answer for 5 and 6 is same

Edited by author 06.01.2021 17:11

One doesn't have to execute their actions followed by the other.
For instance:
Sandro can execute 7 consecutive actions and after that Zagamius could execute 20.
So basically, their turns don't need to be consecutive.
Misinterpretation will lead to WA#4 or WA#5.

Edited by author 05.01.2021 02:08

It is not said in the condition that the points have different coordinates. Can points have the same coordinates?

in some cases, there are more than 1 current user