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Problem link https://codeforces.net/contest/615/problem/D
My Submission https://codeforces.net/contest/615/submission/78009885
Can anyone please tell why my submission is getting WA . I have read the editorial and got the solution but i would like to know why this approach is not working. I am trying this for a long time and have no clue. Please help!!
Auto comment: topic has been updated by saurabh247 (previous revision, new revision, compare).
Look at comments : https://codeforces.net/blog/entry/22658?#comment-271616 and https://codeforces.net/blog/entry/22658?#comment-271626.
you are calculating power(p, k) under modulo 1e9+7 (M). However while calculating k you are taking k = k%M. Instead you have to take k%(M-1) due to fermat's theroem.
For calculating, $$$P = ((cnt[1]+1)*(cnt[2]+1)*(cnt[3]+1)*....*(cnt[k]+1)) * (cnt[i]+1)^{-1} mod (M-1)$$$ , you can precompute : $$$lmul[i] = (cnt[1]+1)*(cnt[2]+1)*(cnt[3]+1)*....*(cnt[i]+1)$$$ $$$rmul[i] = (cnt[i]+1)*(cnt[i+1]+1)*(cnt[i+2]+1)*....*(cnt[k]+1)$$$
Now, $$$P = lmul[i-1] * rmul[i+1]$$$
Now your $$$k = P*((cnt[i]*(cnt[i]+1))/2)mod(M-1)$$$
Thnx but I still have a query .. As you said that for calculating k i have to take mod with (M — 1). I took mod with (M — 1) with my previous submission and got again WA My submission https://codeforces.net/contest/615/submission/78094420 .. I want to know why this is happening ? could u pls help
You are finding inverse of 2 mod (M-1). However it will not exist. Calculate k like this : $$$k=P∗((cnt[i]∗(cnt[i]+1))/2)mod(M−1)$$$
Here before diving by 2, the numerator will be even. So mod (M-1) can be found after dividing by 2.
got it thank you brother I think it's because gcd(2 , 1e9 + 6) is equal to 2.