| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 3845 |
| 2 | jiangly | 3707 |
| 3 | Benq | 3630 |
| 4 | orzdevinwang | 3573 |
| 5 | Geothermal | 3569 |
| 5 | cnnfls_csy | 3569 |
| 7 | jqdai0815 | 3532 |
| 8 | ecnerwala | 3501 |
| 9 | gyh20 | 3447 |
| 10 | Rebelz | 3409 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | maomao90 | 171 |
| 2 | adamant | 163 |
| 3 | awoo | 161 |
| 4 | maroonrk | 152 |
| 4 | nor | 152 |
| 6 | -is-this-fft- | 151 |
| 7 | TheScrasse | 148 |
| 8 | atcoder_official | 146 |
| 9 | Petr | 145 |
| 10 | pajenegod | 144 |
Give an integer N (N<=5000). How many different premutations of sequence 1..n (p[1],p[2]...p[n]) so that p[i]<>i with 1<=i<=n. It means the i-th number in a permutation must be different with its index.
for example
input 2 output 1 (2,1)
input 3 output 2 {(2,3,1); (3,1,2)}
input 4 output 9
Can you give me a solution. thank you very much and sorry about my bad english. :)
from total permutations, subtract number of wrong permutations
Such permutations are called derangements. I believe there is plenty of information about them in the internet.
a[0] = 1
а дальше так: a[n]=(-1)^n + n*a[n-1]
Илиментарная закономерность же, сразу в глаз бросается!!