тест 1: n=9 k=2
тест 2: n=1 k=1
тест 3: n=10 k=1
тест 4: n=8 k=1
тест 5: n=10 k=2
тест 6: n=7 k=2
тест 7: n=1 k=2
тест 8: n=2
тест 9: n=10 k=5
тест 10: n=9 k=5
n=3,4,5,6 не встречается...

incorrectly selected transfer

static void Main(string[] args)
{
   int n = int.Parse(Console.ReadLine());
   int[] values = new int[n];
   for (int i = 0; i < n + 1; i++)
      {
         string[] str = Console.ReadLine().Split();
         int a = int.Parse(str[0]) - 1;
         int b = int.Parse(str[1]) - 1;
         int c = int.Parse(str[2]);

         for (int j = a; j <= b; j++)
            {
               values[j] += c;
            }
      }

   Console.Write(string.Join(" ", values));
}


It's simple algorithm. How to speed it up? Which algorithm should I use?

Edited by author 21.01.2026 21:22

pls some test for W6

Hi, I writed this post because I wanted to share with you some of the test cases* that helped me a lot at the time. I hope it helps you :)

--> Test 1:

1
A B C

--> Answer:

A undefined
B undefined
C undefined

------------------------------

--> Test 2:

5
Isenbaev A B
A B C
D Q P
C H N
G N P

--> Answer:

A 1
B 1
C 2
D 5
G 4
H 3
Isenbaev 0
N 3
P 4
Q 5

------------------------------

--> Test 3:

13
Fominykh Isenbaev BBB
BBB CCC AAA
Ayzenshteyn Oparin Samsonov
Ayzenshteyn Chevdar Samsonov
Dublennykh Fominykh Ivankov
Burmistrov Dublennykh Kurpilyanskiy
Cormen Leiserson Rivest
Oparin AA AAA
Isenbaev Oparin Toropov
AA DD PP
PP QQ RR
RR SS TT
TT Toropov Oparin

--> Answer:

AA 2
AAA 2
Ayzenshteyn 2
BBB 1
Burmistrov 3
CCC 2
Chevdar 3
Cormen undefined
DD 3
Dublennykh 2
Fominykh 1
Isenbaev 0
Ivankov 2
Kurpilyanskiy 3
Leiserson undefined
Oparin 1
PP 3
QQ 4
RR 3
Rivest undefined
SS 3
Samsonov 2
TT 2
Toropov 1

------------------------------

--> Test 4:

3
Isenbaev A B
A B C
A Q W

--> Answer:

A 1
B 1
C 2
Isenbaev 0
Q 2
W 2

------------------------------

*(All test cases were made by other users, so you'll probably found them in other Posts/Topics/Discussions).

Acknowledgments to:
-> "Vit Demidenko" & "Nathalie"
-> "Pavel Nikolov"
-> "mccolt89"
-> "Megakrit"


-How I solved it:

I solved this problem in Java, for that, I used:

->1 java.util.Scanner
->1 java.util.HashMap
->1 java.util.TreeMap
->3 java.util.HashSet
->1 java.util.LinkedList
->? java.util.StringTokenizer (or you can just simply use the Scanner method "next()").

(The structures can be declared inside loops or other structures).

Hope all this helped you :)


Edited by author 08.04.2020 00:48

Edited by author 08.04.2020 00:49

See title I can currently solve problems under a rating of 50(I am a newbie) comfortably but I don't quite get how the difficulty scales really with some problems having a rating of 15000 which seems confusing. I just would like someone who has some knowledge to explain how difficulty scales numerically relative to some standard problems take codeforces or any major competitive programming competition as a reference

Edited by author 08.01.2026 08:49

min(P, 20)

The incorrect test 12 was fixed (coordinates were exceeding 10000).

New tests were added to the problem. The tests target a wide range of problems in geometry algorithms, accuracy issues, and performance.

All solutions have been rejudged. 229 authors (40%) lost, and 2 other authors gained the Solved status for the problem.

I'm expected brute force solution will accepted on C++, because N * 10^D integer divisions might work in 1 sec, but this solution work even on Python3.
Why is work? Is there a math prove that answer will be very small or in this problem weak tests?

Edited by author 14.02.2023 17:29

A bug was fixed in the checker program that incorrectly skipped most of the checks for some tests. The most affected tests are: 1, 3-8, 11, 17.

All solutions have been rejudged. The verdict changed for 40% of solutions (typical change patterns: WA5 -> WA4, WA9 -> WA7, AC -> WA11). 10 authors lost the Solved status for the problem.

how to know when there are no more inputs

#include<stdio.h>
#include<math.h>
int main(){
    unsigned long long int a,b,c,d;
    double Sqrt,Sqrt1,Sqrt2,Sqrt3;
scanf("%llu %llu %llu %llu",&a,&b,&c,&d);
Sqrt=sqrt(a);
Sqrt1=sqrt(b);
Sqrt2=sqrt(c);
Sqrt3=sqrt(d);



printf(" %0.4lf %0.4lf %0.4lf %0.4lf",Sqrt,Sqrt1,Sqrt2,Sqrt3);
}


Edited by author 23.06.2022 13:46

Edited by author 23.06.2022 13:46

A bug fixed in the checker program that allowed passing arbitrary expressions as a first argument to the substring function. According to the problem statements only variables are allowed to be passed there.

Accepted solutions have been rejudged. 26 authors lost the Solved status for the problem.

a , b , c = list(map(int , input().split()))
A , B , C = list(map(int , input().split()))
if a+c >= A:
    if a >= A:
        c =c
        a-=A
    else:
        c=A-a
        a-=A-c
    if b+c >= B:
        if b >= B:
            c =c
            b-=B
        else:
            c=B-b
            a-=A-c
        if a + b + c >= C:
            print('It is a kind of magic')
        else:
            print('There are no miracles in life')
    else:
        print('There are no miracles in life')

else:
    print('There are no miracles in life')

The missing condition added to the English version of the statement:

The cell in which the minus remains should not change.

The Russian version of the statement was correct.

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        String a = "Emperor Penguin";
        String b = "Macaroni Penguin";
        String c = "Little Penguin";
        int countA = 0;
        int countB = 0;
        int countC = 0;
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        List<String> list = new ArrayList<>(n);
        sc.nextLine();
        for (int i = 0; i < n; i++){
            list.add(sc.nextLine());
        }
        for (String item: list){
            if (item.equals(a)){
                countA++;
            } else if (item.equals(b)) {
                countB++;
            } else if (item.equals(c)) {
                countC++;
            }
        }
        int max = Math.max(Math.max(countA,countB),countC);
        if (max == countA){
            System.out.println(a);
        } else if (max == countB) {
            System.out.println(b);
        } else if (max == countC) {
            System.out.println(c);
        }
    }
}

Edited by author 18.11.2017 03:29

If you got WA #28 - check int64 overflow

The limitation on N was changed to reflect the actual test data. Instead of the old N < 10^600 the limitation was set to 0 < N < 10^300.

Note that this is not the limitation on the input, but rather on the output. The number given to the input has about twice as many digits, previously up to ~1200 digits, now up to ~600 digits, but the actual test data was always only up to ~600 digits. Edge case N=0 was also missing in the test data, and is now explicitly disallowed by the limitation.

New thorough tests were added within the new limitations. The time limit was reduced from 2.0 sec to 1.0 sec.

All solutions have been rejudged. 192 authors lost, and 12 other authors gained the Solved status for the problem.