Maybe someone could explain why I'm still getting WA 13? On all of tests in this forum i get right answers. Can anyone give more tests?

Is there any way I could change my user(Author) name?

The triangular sequence is behind the scene.
n belongs to triangular sequence if and only if (8n+1) is a square.

Edited by author 03.02.2021 23:21

Edited by author 03.02.2021 23:21

Edited by author 03.02.2021 23:21

Few lines of code with no conditions

The most time I consumed was to prove that naive algorithm works here :)

4
WMC
WMC
WMC
WMC
---
3
Looking for rogue

4
WM
WMC
WMR
MR
---
4

Okey, he lost the page that had the number of the k-th interesting lamppost.
So, the input contains one integer k.

> Output the number of the k-th interesting lamppost.
Sorry, what? What does input `k` mean then?
____
Я вроде бы родился русским, но никак не могу понять условие задачи. Это что, часть задачи - понять что требуется?

I've used those thing:
- binary search
- sys.stdin.readline() instead of input
- http://acm.timus.ru/forum/thread.aspx?id=30704&upd=636446663176701082
- checked min/max teacher's dates and didn't below/over them

and still have just 1.514-1.544

What's wrong? What's idea behind that?

Сначала пробовал через множества, потом сократил через map
всегда валился на 8 тесте по времени
Решил напрячься
Сел и написал два модуля, чтобы работать с битами
На преподавателе включаю биты
На студенте проверяю и прибавляю к ответу
Сдаю задачу, уже в предвкушении надписи "Accepted", как тут мои глаза лезут на лоб. Снова "TL8"
Что это за монстр такой этот восьмой тест?)

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin>>n;
    while(n--){
        string s;
        int a,b;
        cin>>s;
        a = s[0]-'a'+1;
        b = s[1]-'0';
        a = min(a,9-a);
        b = min(b,9-b);
        if(a==b){
            if(a==1) printf("2\n");
            else if(a==2) printf("4\n");
            else printf("8\n");
        } else {
            if((a==1&&b==2)||(a==2&&b==1)) printf("3\n");
            else if((a==1&&b>=3)||(a>=3&&b==1)) printf("4\n");
            else if((a==2&&b>=3)||(a>=3&&b==2)) printf("6\n");
            else if(a>2&&b>2)printf("8\n");
        }
    }
    return 0;
}

Which is right???

I am stuck with this, it seems impossible to value from input to 2 registers.

What is this test about?

   abc  a
 abc    a

ans
   abc  a

Wrong answer 18. Can anybody help me, please?

В задаче сказано, проверить существует ли такая команда 'x' что она обыграет и команду A и команду В, если существует то ответ No иначе YES

в таком случае можно использовать bitset

Who many test????????????????????

I defined three functions:
C(i) - the number of ways to make the album if it consists of only i songs.
A(i) - the number of ways ending in 1.
B(i) - the number of ways ending in 2.

C(i) = A(i) + B(i)

A(i) = 1 if i <= 1
     = C(i-1) if  1< i <=a
     = C(i-1) - B(i-a-1) i>a

Notice that if i <= a there's no way to have more than a 1's, so we call C(i-1) and 'append' 1 at position i. The number of ways to do this reminds the same as C(i-1).

If i>a you could have an invalid string of 1's. To counter this, we call B(i-a-1) to know exactly how many sequences would be invalid if we 'append' a 1 at position i. When i=a+1, there's only one invalid string resulting of appending 1 at a+1, the sequence consisting of only 1's.

B is analogous to A (calls A instead of B and uses b instead of a).

Edited by author 12.07.2019 04:31