### maroonrk's blog

By maroonrk, history, 7 weeks ago, We will hold AtCoder Regular Contest 167.

The point values will be 300-500-700-700-800-1200.

We are looking forward to your participation! Comments (20)
 » Why is F given 1200 score...
 » Wish I won't drop back to 1 Kyu.
 » emm...
 » Stuck at C for more than 45 minutes before trying D and realizing that it is pretty doable.
 » Is there any O(T) solution for E?
•  » » Oh I found such a submission. I only thought of solutions for S=2k,3k,3k+2, but there's a simple solution for S=2k+1.
•  » » In fact only one in-contest submission used extended gcd while the other 12 are $O(T)$
 » my last formula of my sol is similar to editorial >>why it give me wa on B my solution `long long a, b; cin>>a>>b; vectorphi; for(long long i = 2;i*i<=a;i++){ if(a%i==0){ long long cnt = 0; while(a%i==0) cnt++, a/=i; phi.push_back(b*cnt+1); } } if(a>1) phi.push_back(b+1); long long ans = 1; for(auto i:phi){ ans = ans * (i%md); ans%=md; } ans = ans * (b%md); ans%=md; ans = ans * fst_pow(2, md-2); ans%=md; cout<
•  » » Because ans may not be even before you divide by 2. Correct answer is floor(ans/2).
•  » » » 7 weeks ago, # ^ | ← Rev. 4 →   okay thanks>>I changed the mod to 2 * mod and divide by 2 in the last and did not use 2^mod-2
 » can anyone explain about this? It seems that there exict a O(1) solution of problem E. https://atcoder.jp/contests/arc167/submissions/46638239
 » Can anyone get me clear on the concept behind B? i did not get the editorial
 » Can anyone please explain why is my submission getting TLE for some cases where as it is running under 2 ms for all other cases.I'm doing exactly what editorial says : B * product of (e_i * B + 1) for all powers of prime factors, finally dividing it by 2.
•  » » Overflow probably
•  » » » Yes, indeed it is. for loop condition : j * j <= A got overflowed thus making it endless loop.Thanks a lot for the help!!
 » why my solution fails?: problem Bmy approach: we gonna factorize A and focus on one prime from the factorization: I need to know how mcuh p there is in the product of the divisors of a^b suppose a=p^k * .. so the number of this prime in the product is: (sum from 1 to k*b)* product of (ki +1) where ki is the power of the otehr primes sequencially.after that I just divide by k why this fails? : https://atcoder.jp/contests/arc167/submissions/46638531
•  » » It's wrong. You don't divide by k, you multiple by B then divide by 2. You also have to make sure if the number is odd before dividing by 2. This is doable by checking if all the numbers you're multiplying are odd
 » Can someone explain solution of problem C?
 » 7 weeks ago, # | ← Rev. 2 →   I think problem E would be a good math problem, but I don't think it's a good CP problem. It's so tricky that you can find 4 people who's rating is under 1600 among 13 people who solve the problem during the contest:(
 » About Problem B :Can anyone tell me why I set the mod 998244353 the answer was WA,but when I changed it to 998244353*2 ,it was AC ?