The problems will be as difficult as the august long challenge?
It can be, it may not be, cannot be sure, first 3-4 are simple-easy then it heats up
In DIV-2, first 5 problems(and equivalently 2 in DIV-1) are quite easy, it is the 6th problem from where the fun begins.
what does this mean ?
The first user to solve the problems in the Long Challenge(except the Tiebreaker problem), will get 100 laddus. This is what the statement means.
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Hoping for something exciting 
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Very much excited! Let’s do this. _/_
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Let’s suppose that you are the first one in solving k problems (no other contestant got full score before you in any of those problems), then you will get 100*k laddus;
Got it Thankyou. …
Unfortunatly it is lot of math this time.
In fact, I would call only two of the A problems “programming problem”, the others are more like “math problems”.
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This is not true at all 
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Hello, Editorials for most problems are ready , and are in verification phase. They will be released at the earliest
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By looking at the small constraints, brute force (dp) was the only thing that popped in my mind. 
I did figure out the simpler approach later…
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No need to use dp, You should check out Very Easy Solution here for Chef and Interesting Subsequences
from sys import stdin, stdout
import math
def main():
t = int(stdin.readline())
for _ in range(t):
n, k = map(int, stdin.readline().split())
a = list(map(int, stdin.readline().split()))
a.sort()
mx = a[k - 1]
nn, r = 0, 0
for i in range(k):
if a[i] == mx:
r += 1
for i in range(k, n):
if (a[i] == mx):
nn += 1
else:
break
nn += r
ans = math.factorial(nn) / (math.factorial(nn - r) * math.factorial(r))
print(int(ans))
if __name__ == "__main__":
main()
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Any ETA on Tomorrow night for sure my friend .
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I think N \le 100 should work, but anything beyond that won’t - the people who hate everyone else cause my solution to explode 