I could solve it using trivial bruteforce. But I get TL, using more clever and probably expected solution. I used this: https://e-maxx.ru/algo/discrete_root (Read only first section, we don't need anything else). The slowest part here is Discrete Logarithm, which I calculate in \sqrt{n} * log_from_map.

My program stuck at WA-17 for several times and I don't know the reason why. Can anyone give me some challenging test cases?

The more - the better! 'cause I have no idea what I'm doing wrong :\

#include <iostream>
using namespace std;

int main() {
    int leavingCars, minutes;
    int carPerMinute[100];
    int totalCars = 0;

    cin >> leavingCars >> minutes;
    int carsLeavingPerMinute = leavingCars * minutes;

    for(int index = 1; index <= minutes; ++index) {
        cin >> carPerMinute[index];
        totalCars += carPerMinute[index];
    }

    cout << totalCars - carsLeavingPerMinute;

    return 0;
}

Are you going to add javascript (especially nodejs) support for this online judge | Планируется ли добавить на сайт поддержку javascript или nodejs?

not 21 passes test:

using System;
using System.Collections.Generic;
using System.Globalization;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApplication3
{
    class Program
    {
        static void Main()
        {
            var f = Console.In.ReadToEnd();
            f = f.ToLower();
            var b = f.ToCharArray();
            string c = "";
            c += b[0].ToString().ToUpper();
            for (var i = 1; i < b.Length - 3; i++)
            {
                if ((i + 1 != b.Length && (b[i] == '!' || b[i] == '?' || b[i] == '.'||b[i] == '-') && (b[i+3] != '-'&&b[i + 1] != '!' && b[i + 1] != '?' && b[i + 1] != '.')))

                {
                    c += b[i];
                    i++;
                    while (b[i] == ' ' || b[i] == '\t' || b[i] == '\r' || b[i] == '\n')
                    {
                        c += b[i];
                        if (i + 1 != b.Length)
                            i++;
                        else break;
                    }
                    c += b[i].ToString().ToUpper();
                }
                else
                    c += b[i];
            }


            Console.WriteLine(c);
        }
    }
}



not 4 passes test:

using System;
using System.Collections.Generic;
using System.Globalization;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApplication3
{
    class Program
    {
        static void Main()
        {
            var f = Console.In.ReadToEnd();
            f = f.ToLower();
            var b = f.ToCharArray();
            string c = "";
            for (var i = 1; i < b.Length - 1; i++)
            {
                if (b[i - 1] == '-')
                    i--;
                if ((i + 1 != b.Length && i + 2 != b.Length && i + 3 != b.Length && (b[i] == '!' || b[i] == '?' || b[i] == '.' || b[i] == '-') && (b[i + 3] != '-' && b[i + 1] != '!' && b[i + 1] != '?' && b[i + 1] != '.')))
                {

                    c += b[i];
                    i++;
                    while (b[i] == ' ' || b[i] == '\t' || b[i] == '\r' || b[i] == '\n')
                    {
                        c += b[i];
                        if (i + 1 != b.Length)
                            i++;
                        else break;
                    }
                    c += b[i].ToString().ToUpper();
                }
                else
                {
                    if (i == 1)
                        c += b[i - 1].ToString().ToUpper();

                    c += b[i];
                }
            }


            Console.WriteLine(c);
        }
    }
}

Edited by author 03.01.2015 21:08

Here are jokes with statuses, such as we go in order along the vertices and if we get to a vertex that we haven't been to, then we start going in a cycle from this vertex, setting the status of this cycle, if we get to the same status, then this is a cycle, if we get to a vertex with a lower status, then everything is fine and we stop the cycle, and if the vertex has no status, then we just opened a new vertex and it turns out that the algorithm is linear

I've been trying to solve this problem for so long, I've had a lot of errors, more than 150 lines of clean code, I've tried to do a little dirty debug, but I think it's not fair. I got wa11, I didn't know what it was, it just turns out I forgot to sort the array correctly, it was in the second part when I'm looking for vertical stripes that I forgot to sort and got wa 11, while I was looking for it, I did a lot of tests and can show them to you. I can give advice on what to count, count all vertical rectangles 1*l where l>=2 and horizontal rectangles l*1 where l>=2, add them up and add all single cells.
tests:
4 4 10
1 2
1 3
2 1
2 4
3 2
3 3
4 1
4 2
4 3
4 4
5

4 4 5
1 2
2 1
2 4
3 3
4 2
9

4 4 8
2 1
2 2
2 3
2 4
3 1
3 4
4 2
4 3
4

4 4 8
1 2
1 3
2 1
2 3
3 1
3 3
3 4
4 2
5

4 4 4
2 2
2 3
3 2
3 3
4

4 4 3
2 3
3 2
3 3
6

5 5 2
4 3
3 4
12

5 5 5
1 3
3 1
4 1
3 4
4 3
12

5 5 6
1 3
3 1
4 1
4 2
4 3
3 4
12

5 5 12
1 2
1 4
2 1
2 3
2 5
3 2
3 4
4 1
4 3
4 5
5 2
5 4
13

5 5 11
1 2
1 4
2 1
2 3
2 5
3 2
3 4
4 1
4 5
5 2
5 4
11

5 5 9
1 2
2 4
3 1
3 3
3 4
4 2
4 5
5 2
5 5
11

5 5
9
1 5
2 5
2 4
3 3
3 4
4 2
4 3
5 1
5 2
10

5 5 12
1 2
1 3
1 4
2 1
3 1
4 1
2 5
3 5
4 5
5 2
5 3
5 4
10

6 6 12
2 1
2 2
2 3
2 4
2 5
3 5
4 5
5 5
5 4
5 3
5 2
4 2
8

7 4 14
1 1
2 1
3 1
4 1
5 1
6 1
7 1
1 4
2 4
3 4
4 4
5 4
6 4
7 4
9

7 4 18
1 1
2 1
3 1
4 1
5 1
6 1
7 1
1 4
2 4
3 4
4 4
5 4
6 4
7 4
1 2
1 3
7 2
7 3
7

7 4 20
1 1
2 1
3 1
4 1
5 1
6 1
7 1
1 4
2 4
3 4
4 4
5 4
6 4
7 4
1 2
1 3
7 2
7 3
4 2
4 3
8

7 4 28
1 1
2 1
3 1
4 1
5 1
6 1
7 1
1 4
2 4
3 4
4 4
5 4
6 4
7 4
1 2
1 3
7 2
7 3
4 2
4 3
3 2
3 3
2 2
2 3
5 2
5 3
6 2
6 3
0

7 4 27
1 1
2 1
3 1
4 1
5 1
6 1
7 1
1 4
2 4
3 4
4 4
5 4
6 4
7 4
1 2
1 3
7 2
4 2
4 3
3 2
3 3
2 2
2 3
5 2
5 3
6 2
6 3
1

7 4 7
1 2
2 3
3 2
4 3
5 2
6 3
7 2
9

7 4 1
3 2
12

7 5
1
2 3
13

7 5 2
3 3
4 3
15

7 5 10
1 2
2 2
5 2
6 2
7 2
1 4
2 4
5 4
6 4
7 4
7

7 4 14
1 1
3 1
5 1
7 1
2 2
4 2
6 2
1 3
3 3
5 3
7 3
2 4
4 4
6 4
14

5 1 4
1 1
3 1
4 1
5 1
2
WRONG!!!
on this test i have eror, right ans is 1
good luck, don't give up it's not such a difficult problem!

Каким образом во втором тесте(примере) получаем 2 полосы???
если даже когда вручную нарисовав таблицу 1х5, получаем максимум 1 полоску????каким образом вторая получается???
п.с.или белая полоса может состоять из одного дня????

My program stucked at WA-25 for several times and I don't know the reason why. One thing that confuses me is that I don't know how to deal with the following sentences in problem statement:
"You should consider a rotation to map a key point A to a key point B if A is mapped to a point within 10−6 units distance from the point B. Rotations having the same mapping should be considered as the same rotation."
I think the second sentence is redundant since a rotation is uniquely determined by its images on two different non diametral points. I don't know how to implement a function such that, given the images q_i of each point p_i, determine if any rotation f exists such that dist(f(p_i), q_i)<1e-6.

I had TL 5 because I did not add vertices to the BFS queue properly (sometimes I put the same vertex multiple times and then process in many times). You may try this test:

[test cut, too big]

Edited by moderator 06.09.2022 03:04

Firstly, I pre-calculate the index which carries 1, and store them on the map. then I take input and check the map if the map stored 1 for this input, then I printed 1, otherwise 0.
what is wrong here?

I walked through all topics in discussion, grabbed every test, but still got WA7.

I had several mistakes in my solution.

First of all, in some versions I forgot to make fixed format of output and forgot to cast my answer to integer type, therefore the solution printed the answer in scientific notation.

Test:
25
7217 6651
7358 2764
7034 5638
2431 6831
6550 2019
2334 1524
3375 4802
5960 4579
7078 1798
3176 7335
5872 3950
179 7351
9800 3982
2080 897
9345 2389
6470 6556
8420 9316
1671 7557
922 2613
9464 6735
1865 8925
6701 3365
6519 6665
9458 1535
8792 4383
Answer:
1495422

In some solutions there were rounding issues, I rounded down while should have rounded up.
Still I disagree that sometimes one should use floor(ans) instead of int(round(ans)), the latter always works fine in this problem.

Test:
3
4826 9656
137 7424
8340 2686
Answer:
22472

In some attempts I accidentally used = instead of ==, this test detected this perfectly.

Test:
4
0 0
2 0
2 2
0 2
Answer:
14


Edited by author 01.09.2022 22:29

ITS REALLY HARD
I WORK SO HARD FOR THIS PROBLEM TO DO ITS REALLY FAST, U HAVE 1MB MEMORY input THIS IS ABOUT 600-700k symbols, U NEED TO DO ITS Very VERY VERY VERYVERY VERy fAST AND THIS VERYV VEryV eVEYR Very HARD
U NEED TO READ THIS NEVER NEVER NEVER USE CIN
USE c=getchar() where c - char. And IF U WANT TO STOP INPUT JUST int WHILE
in the conditions of the while loop, put restrictions on the symbol, that is, there are all charms from A to Z and something else. AND THEN YOU NEED A REALLY FAST ALGORITHM
I rewrote mine six times, I can't understand it myself, it's very difficult, I only remember that char is int, and int is a deadline, and diffirintly THE ALGORITHM is SLOW

Please remember that input can contain numbers with more than one digit. My WA was due to this, and I don't think anybody will be this silly. But still posting! :)

Edited by author 29.08.2022 21:23

Edited by author 29.08.2022 21:23

delete iostream and using namespace std.
all cin cout replace on scanf and pritf
write in start of code: #define CRT_SECURE_NO_WARNINGS
and #include<cstdio>
and behind your vectors,pair and queue write std::
do vector WITHOUT ANY INTS, ALL BOOL
DONT USE PUSH_BACK its really slow, just write size of vector
Example:
>std::vector<std::vector<bool>> a(n);
>std::vector<bool> b(m);
and call with a[i][j] or something else;
If you look at other aspects of the task, then EASY BFS

String can have size 1000, max size is not 999.