36 60 60238766947 689134144379425669 151303190236896619 105037406971508485 68317121395
x=6221392564

#include<iostream>

using namespace std;
int main()
{
int a;
int b;

int c;
int d;
int h;
cin>>a;
cin>>b;
h=a-1+b;
c=h-a;
d=h-b;
cout<<c;
cout<<d;

return 0;
}

Hello,

I have AC now. I used code from here:
http://stackoverflow.com/a/21032389

Is there more clever way to value (a*b)%c?

I don't understand problem,please give me some tests.Thank!!!!!!

let there be 2k odd gaps and n-2k even gaps between island A
even gaps could be BCBCBC or CBCBCB
odd gaps are half BCB half CBC

Try this test:
===========================
-10 -10
10 5
5 10
0 0
---
42.8571428571
5

Edited by author 11.05.2021 19:41

With C++: code: https://pastebin.com/gJu5U454 I get WA

The same code with Python is Accepted : https://pastebin.com/Cei31LkY

C++ I used unsigned long long !!!!!!!!!!!!!

I checked the output of my code for all K ( 1 <= k <= 65535), I think it gives the correct output.
https://github.com/Mourad-NOUAILI/Timus-Online-Judge/blob/main/1209/WA%233

long long type is used.

Code explanation:
I use a(k) = 1, 2, 2, 4, 4, 4, 8, 8, 8, 8, 16, 16, 16, 16, 16, 32, 32, 32, 32, 32, 32, 64, 64, 64, 64, 64, 64, 64,... to compute a(k)
https://oeis.org/search?q=1%2C2%2C2%2C4%2C4%2C4%2C8%2C8%2C8%2C8%2C16%2C16%2C16%2C16%2C16&sort=&language=english&go=Search

the pos(k) in a(k) from 0, 1, 0, 2, 1, 0, 3, 2, 1, 0, 4, 3, 2, 1, 0, 5, 4, 3, 2, 1, 0, 6, 5, 4, 3, 2, 1, 0, 7, 6, 5, 4, 3, 2, 1, 0, 8, 7, 6, 5, 4, 3, 2, 1,..
https://oeis.org/search?q=1+0+2+1+0+3+2+1+0+4+3+2+1+0&sort=&language=english&go=Search

sequences are 0-base indexing.

example:
k = 16 --> a(15) = 32 = (100000)2

pos(15) in 0, 1, 0, 2, 1, 0, 3, 2, 1, 0, 4, 3, 2, 1, 0, 5, 4,... is 5, which means we gonna pick the bit at the 5th position from the left in 100000

100000
543210

so, for k = 16, the answer is 1

Thanks.

Edited by author 21.05.2021 00:09

import math
import operator

n = int(input())

s = input()
l1 = s.split(" ")
x = int(l1[0])
y = int(l1[1])

sp = {}


for i in range(n-1):
    s = input().split(" ")
    x1 = int(s[0])
    y1 = int(s[1])
    dx = x1-x
    dy = y1-y
    dl = math.sqrt(math.pow(dx,2)+math.pow(dy,2))
    a = math.degrees(math.asin(dy/dl))
    #print(a)
    sp.update({i+2:a})

sorted_tuples = sorted(sp.items(), key=operator.itemgetter(1))
#print(sorted_tuples)
print("1")
print(sorted_tuples[len(sorted_tuples)//2][0])

2 1
2
Ответ: 1
Этот тест может быть крайним случаем в вашем решении так как уже на такой тест и схожие:
3 2
2
3
Ответ: 2

Edited by author 05.05.2021 06:45

Я получил такой вердикт, выводя "Ask Shiftman for help." как только встречалось в цикле подходящее условие, вместо того чтобы проверить все отрезки до конца на наличие случая "Let's search for another office."

m=[]
m2=[]

for tik in range(3):
    n=int(input())
    m1=input().split()
    for i in m1:
        if i not in m:
            m.append(i)
        elif i in m:
            if i not in m2:
                m2.append(i)

print(len(m)-len(m2))


I don't know why this program doesn't work.

Edited by author 03.05.2021 00:22

If you put two books on a bookshelf by facing binding side (book number) outwards i.e facing person, then this problem statement makes sense.

The coordinates are given in the traversal order.

Try this:
3 3 6
1 1
1 2
1 3
2 1
2 2
2 3


Should be:
1

input:
5 5
1 2 0
2 3 1
2 4 1
3 4 1
4 5 2

one possible output:
1 2 3 5

wrong output:
1 2 3 4 5

Don't forget that the output also needs to have as few edges as possible!