№ | Пользователь | Рейтинг |
---|---|---|
1 | jiangly | 3640 |
2 | Benq | 3593 |
3 | tourist | 3572 |
4 | orzdevinwang | 3561 |
5 | cnnfls_csy | 3539 |
6 | ecnerwala | 3534 |
7 | Radewoosh | 3532 |
8 | gyh20 | 3447 |
9 | Rebelz | 3409 |
10 | Geothermal | 3408 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | maomao90 | 173 |
2 | adamant | 164 |
3 | awoo | 161 |
4 | TheScrasse | 160 |
5 | nor | 159 |
6 | maroonrk | 156 |
7 | SecondThread | 152 |
8 | pajenegod | 146 |
9 | BledDest | 144 |
10 | Um_nik | 143 |
Название |
---|
the input size if small (the largest factorial you will need is 20!) so here you can generate all the possible factorials and use bitmasking (or recursion) to pre-calculate all possible values and cache them ... then for every case you just output the answer
I pre-calculate all the value upto 20. But now i cant understand how can i find my desired number .
I have no idea about bitmasking. Is there any other way ?
Use brute-force. For each number there are two cases — to use or not to use :)
If i use brute-force, at some stage summation of fact(num) may exceed my desired n , at that case i may avoid that fact(num) . But can i surely say that at any stage summation of fact(num) will be equal to my desired n ?
I think that this approach will result in TLE since the limit is 0.5 :) I think it is better to read Caraz96's idea and ask if you misunderstood something in it.
It is possible to use a greedy approach: since fact(n) > fact(0)+fact(1)+...+fact(n-1) (this is true for each n > 2) you can iterate from the greatest factorial(20!) to the smallest one, if fact[i] is less than or equal to N you must take it, decrease N by fact[i] and proceed with the next factorial. At the end, if N is 0 you found the solution, otherwise it is impossible to obtain N adding factorials.