For Binomial Fever problem I used simple geometric progression for: Test Case r=2: my code is ‘’’ for r=2 ans = pC2+p^2C2+… ans = (p(p-1)+p^2(p^2-1)+…+p^n(p^n-1)/2 ans = ((p^2 + p^4 + p^6 + … + p^2n) - (p + p^2 + p^3 +…+ p^n))/2 ans = (G.P of p^2 - G.P of n)/2 ‘’’ t = int(input()) m = 9…

I’ve used the same logic, and get WA in SB 2

For all cases where p=998244352, ie p=mod-1, inverse modulo does not exist for (p^2-1) as one of its factor is (p+1) , which is mod itself. As you know for inverse modulo to exist , the number should be co-prime with the modulo . Ex: 1 998244352 2 Your code will give wrong answer (0) for this cas…

Thank you @njha1999 Now i got it

Cool! I was stuck at the same problem but how do you get around it? A normal for loop would take 10 sec to calculate 10e9 998244352 2. I temporarily got around it by using an if condition assuming the above test case( and i was right lol)

and yes simply if p == mod-1 and r =2 then print(ceil(n/2) … @singhpriyank (for 15 marks see —> CodeChef: Practical coding for everyone )

basically i saw a pattern from n=1 to100 ( when p ==mod-1 and r =2) the answer should be 1 1 2 2 3 3 and so on but when i saw my solution gives 0,0,0,0,… so i write a condition if p==mod- 1 and r =2 then print(ceil(n/2)) (for 15 marks see —> CodeChef: Practical coding for everyone )

(P/Q) %m, when gcd(Q,m)!=1, is ((P % (Q*m)) / Q) This will be sufficient for 15 pts in python, you may need big int in c++ for using this. For other subtasks: Sum of GP by recursion will help you: 1 + a + a^2 + ... + a^(2*n+1) = (1 + a) * (1 + (a^2) + (a^2)^2 + ... + (a^2)^n) Use this to get su…

Along with this i used divide and conquer over NTT(not my implementation :frowning: ) to get 100 pts.

Binomial Fever (BINOFEV)

help

just1star December 16, 2019, 11:16am 14

Subtask 4 of this editorial can help I guess.

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**Binomial Fever (BINOFEV)**

Binomial Fever (BINOFEV)