"lesser the age gap between two people, higher the interaction between them" so we need to find the first neighbour pair with the lowest age gap

n=int(input())
A=[int(x) for x in input().split()]
 
from itertools import pairwise
req=min(abs(x-y) for x,y in pairwise(A))
 
for i in range(n-1):
    if abs(A[i]-A[i+1])==req:
        print(i+1)
        break

he gets the verdict "Wrong answer on test k"

So, he knows the correct answer for the first k-1 test cases and also he knows the wrong answer for the test case k. The opposite of his current answer for test case k will be the correct answer. Hence, he now knows the correct answer for the first k test cases

There are m-k test cases remaining. In the worst case scenario, all of his initial guesses would be incorrect. Hence he will get m-k more WAs

He initially got a WA. So, in total he got m-k+1 WA

m,k=map(int,input().split())
print(m-k+1)

TLDR: the edge case is 2028

• If x%4=1, that year's batch will be assigned hostel A

• If x%4=2, that year's batch will be assigned hostel B

• If x%4=3, that year's batch will be assigned hostel C

• If x%4=0, that year's batch will be assigned hostel D

Since there are 50 students per floor,

• In 2025 ground floor and 1st floor occupied (2 floors in total)

• In 2026 ground floor to 2nd floor occupied (3 floors in total)

• In 2027 ground floor to 3rd floor occupied (4 floors in total)

• In year x, ground floor to (x-2024)th floor occupied (x-2023 floors in total)

If x%4=1, first floor is guaranteed to be occupied. So answer will be 1/(x-2023)

If x%4=2, ground floor is guaranteed to be occupied. So answer will be 1/(x-2023)

If x%4=3, ground and first floor are both guaranteed to be occupied. So answer will be 2/(x-2023)

If x%4=0 and x>2028, first, third and fifth floors are guaranteed to be occupied. So answer will be 3/(x-2023)

If x%4=0 and x=2028, fifth floor will not be occupied but first and third floors will be occupied. So, answer will be 2/5

Kindly note the given output format in the problem

from math import gcd
for _ in range(1):
   x=int(input())
   if x==2028:
       print("2 5")
       continue
   b=x-2023
   if x%4==1:
       a=1
   elif x%4==2:
       a=1
   elif x%4==3:
       a=2
   else:
       a=3
   g=gcd(a,b)
   a//=g
   b//=g
   print(a,b)

By replacing exactly one bead, we can create one beautiful necklace

For the cases where we should replace exactly 2 beads, first let's pick an arbitrary white bead and replace it with a red bead. Now, two configurations will be different iff the minimum distance between the two beads are different in the two configurations.

• If n is even, the first choosen bead has an opposite bead. we can start from either neighbours of the initial bead and go step by step till that opposite bead. Each of them will represent a distinct configuration since the minimum distance is just the path traced. So, there will be n/2 possible configurations with 2 beads being replaced

• If n is odd, there is no opposite bead. But now, if we cut the necklace at the string opposite to the initial bead, there are exactly (n-1)/2 beads on either sides of the initial bead and they're all guaranteed to be the minimum distance. Hence, the answer will be (n-1)/2

So, the final answer will be 1+(n/2) if n is even else 1+(n-1)/2

for _ in range(int(input())):
    n=int(input())
    print(1+n//2)