| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 3993 |
| 2 | jiangly | 3743 |
| 3 | orzdevinwang | 3707 |
| 4 | Radewoosh | 3627 |
| 5 | jqdai0815 | 3620 |
| 6 | Benq | 3564 |
| 7 | Kevin114514 | 3443 |
| 8 | ksun48 | 3434 |
| 9 | Rewinding | 3397 |
| 10 | Um_nik | 3396 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | cry | 167 |
| 2 | Um_nik | 163 |
| 3 | maomao90 | 162 |
| 3 | atcoder_official | 162 |
| 5 | adamant | 159 |
| 6 | -is-this-fft- | 158 |
| 7 | awoo | 156 |
| 8 | TheScrasse | 154 |
| 9 | Dominater069 | 153 |
| 9 | nor | 153 |
Problem link https://mirror.codeforces.com/contest/615/problem/D
My Submission https://mirror.codeforces.com/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://mirror.codeforces.com/blog/entry/22658?#comment-271616 and https://mirror.codeforces.com/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://mirror.codeforces.com/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.