| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 3757 |
| 2 | jiangly | 3647 |
| 3 | Benq | 3581 |
| 4 | orzdevinwang | 3570 |
| 5 | Geothermal | 3569 |
| 5 | cnnfls_csy | 3569 |
| 7 | Radewoosh | 3509 |
| 8 | ecnerwala | 3486 |
| 9 | jqdai0815 | 3474 |
| 10 | gyh20 | 3447 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | maomao90 | 171 |
| 2 | adamant | 164 |
| 3 | awoo | 163 |
| 4 | TheScrasse | 159 |
| 5 | nor | 155 |
| 6 | maroonrk | 154 |
| 7 | -is-this-fft- | 152 |
| 8 | Petr | 147 |
| 9 | orz | 145 |
| 9 | pajenegod | 145 |
Link: https://uva.onlinejudge.org/external/108/10883.pdf
I could figure out that let's say for 5 numbers the answer is (Arr[1] + 4 * Arr[2] + 6 * Arr[3] + 4 * Arr[4] + Arr[5])/2^16. So the problem basically reduces to finding the values in the nth row of pascal's triangle and power of 2. But I am not very sure as to how to implement the solution. Any help is really appreciated.
HINT: If you calculate in the paper, you will see answer is
You can use the fact that
, i.e. no need of pre-computation. You can find a similar problem here
Yeah, recursively like C(n, k) = C(n, k - 1) * (n - k - 1) / k
HINT 2: Number of combinations can calculated recursively: C(n, k) = C(n, k - 1) * (n - k - 1) / k
Also C(n, 0) = 1
HINT 3: For bigger n finding combination will not be work, so use logarithm : )
If you couldn't solve, see solution with logarithm and numerical combinations