It says we have to start at the lowest left tile. Does that mean that the euclidean distance of the tile from origin should be minimum?? or does it mean that we have to look for left most layer and lowest tile in that??

try this test:

label: a = 1
goto label

I try to solve this problem as "offline LCA problem" with recursive DFS, but I have TLE? Where is mistake?

P.S. Нашёл баг, из-за которого был WA: если вы в Pascal пишите z:=n*(n-1), где z-int64, n-longint, то результат z будет типом longint; правильно так: z:=n; z:=z*(n-1).
И ещё: лучше выводить не 4 знака в ответе, а все.

Edited by author 03.08.2016 11:44

KMP + DP, O(n ^ 2)

2
87 4547
87 2945 1516
64 913 9864
-----------
9904.472185221

A solution that doesn't make use of the max number being 10,000 is using inclusion-exclusion principle. Because when we take the sets of K that have a gcd divisible by 2 and the sets of K that have a gcd divisible by 3, you count twice the ones with a gcd divisible by 6, so you have to subtract those. It's similar to a solution to this problem: https://open.kattis.com/problems/coprimeintegers

import java.util.*;
public class j {
    public static void main(String[] args){
        Scanner input = new Scanner(System.in);
        int bin,sum = 0;
        int n=input.nextInt();
        int m=input.nextInt();
        int [] mas = new int [n];
        for(int i = 0;i<m;i++){
            bin = input.nextInt();
            sum += 1;
            mas[bin-1] = mas[bin-1] + 1;

        }
        for(int i = 0;i<n;i++){
            System.out.println(String.format("%.2f",mas[i]/(sum*0.01))+"%");
        }

    }
}

import java.util.*;
public class Main {

    public static void main(String[] args) {
        ArrayList<Integer> mas = new ArrayList<>();
        Scanner input = new Scanner(System.in);
        int n,bin,delta = 0;

        n = input.nextInt();
        for (int i = 0; i <n;i++){
            bin = input.nextInt();
            mas.add(bin);
        }

        n = input.nextInt();
        for (int i = 0; i <n;i++){
            bin = input.nextInt();
            mas.add(bin);
        }

        n = input.nextInt();
        for (int i = 0; i <n;i++){
            bin = input.nextInt();
            mas.add(bin);
        }

        Collections.sort(mas);
        delta = mas.size();
        for (int i = 0;i<mas.size()-1;i++){
            bin = mas.get(i);
            delta = delta - (mas.lastIndexOf(bin)-mas.indexOf(bin)+1);


        }
        delta = -1*delta-mas.size();
        System.out.print(delta);
    }
}

:P

import java.util.*;

public class ReverseRoot {
    public static void main(String[] args) {
        ArrayList<Long> list = new ArrayList<>();
        Scanner scn = new Scanner(System.in);
        while (scn.hasNextLong()) {
            long p = scn.nextLong();
            list.add(p);
        }
        scn.close();
        for (int i = list.size() - 1; i >= 0; i--) {
            System.out.printf("%.4f%n", Math.sqrt((double) list.get(i)));
        }
    }
}

Does anyone know about test number 5? If you do, please download the values for this test.

write line "#define int long long" and change "int main" to "signed main"

What test number 2 can be?

Edited by author 21.02.2021 00:09

program p1071;
var x,y,z,s,i:longint;
    out:boolean;
procedure check(p:longint);
var  x1,y1:longint;
     c:array[0..10000] of integer;
begin
     fillchar(c,sizeof(c),0);
     x1:=x;y1:=y;
     repeat
           inc(c[x1 mod p]);
           x1:=x1 div p;
     until x1=0;
     repeat
           if c[y1 mod p]>0 then dec(c[y1 mod p])
                            else exit;
           y1:=y1 div p;
     until y1=0;
     out:=true;
end;
begin
     readln(x,y);z:=0;out:=false;
     for i:=2 to x do
         if x mod i=y mod i then
         begin
              check(i);
              if out then break;
         end;
     if out then write(i)
            else write('No solution');
     readln;
end.

Does anyone have a clue on what test №16 is? If something, I am writing on Python.

Edited by author 19.02.2021 02:12

it wasted my whole day using string.
then i use array instead of boring string and accepted

a valid n digit number can be found from valid n-1 and n-2 digit numbers

Edited by author 16.02.2021 13:56

Edited by author 16.02.2021 13:56

[DELETED]

Edited by author 17.09.2022 01:09

Easy solution with O(p) and without using a primality of p. We can build a table of square roots modulo p and then...