You only have four things to pick from, so you can calculate each step explicitly. This equation follows the notation in the article: T is the time to collect all of the items, and ti be the time to collect the i-th item after i−1 items have been collected.

E(T)=E(t1)+E(t2)+E(t3)+E(t4)=p1−1+p2−1+p3−1+p4−1.
The probability p1 of picking a new one if you have picked none yet is 1.

The probability p2 of picking a new one if you have picked one depends on which one you picked first. Let's call the items A through D and the probabilities of picking each item if they're all in the box pa through pd.

Then p2 would be
pa(pb+pc+pd)+pb(pc+pd+pa)+pc(pd+pa+pb)+pd(pa+pb+pc).
The above expression is the probability of picking a new one given that A had been picked already, plus the probability of picking a new one given that B had been picked already, and so on.

The expression for p3 will have six terms in the sum; one of these will be papb(pc+pd). The expression for p4 will have four terms in the sum (actually, it's the same term four times!)

What can I do?.. My program passes all testes here, but still WA12... Any test cases?

it takes a lot of my times.

#include<iostream>
using namespace std;
int main()
{
    int a, b;
    cin >> a >> b;
    cout << b-1 << ' ' << a - 1;
}

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

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

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

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

diameter of 6875 miles (/ 2 = 3437.5)???

the mean radius of earth is 6371.0 km (3958.8 miles)

a = input()
minussum=0
plussum=0
b =[0:3]
c =[3:6]
c=int(c)
minus = c-1
plus = c+1
minus=str(minus)
plus=str(plus)
for i in minus:
    i=int(i)
    minussum+=i
for i in plus:
    i=int(i)
    plussum+=i
n=0
for i in b:
    i=int(i)
    n+=i
if n == minussum or n == plussum:
    print("Yes")
else:
    print("No")




Help please)
Why re?

Edited by author 13.09.2022 23:21

Python files encoded in UTF-8 with BOM always get "Runtime Error 1".
It would be nice if it was fixed.

I just do maxsimum 30 iteration on my super all doing bruteforce and this give me right ans and i got ac

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????