the question is very simple we just need to calculate total number of numbers which have exactly 4 divisors for ex 6, 8, 10 these are all of the forms p^3 or p*q but here n<=10^11
the question is very simple we just need to calculate total number of numbers which have exactly 4 divisors for ex 6, 8, 10 these are all of the forms p^3 or p*q but here n<=10^11
№ | Пользователь | Рейтинг |
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1 | tourist | 3843 |
2 | jiangly | 3705 |
3 | Benq | 3628 |
4 | orzdevinwang | 3571 |
5 | Geothermal | 3569 |
5 | cnnfls_csy | 3569 |
7 | jqdai0815 | 3530 |
8 | ecnerwala | 3499 |
9 | gyh20 | 3447 |
10 | Rebelz | 3409 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
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1 | maomao90 | 171 |
2 | awoo | 163 |
3 | adamant | 162 |
4 | TheScrasse | 158 |
5 | nor | 153 |
5 | maroonrk | 153 |
7 | -is-this-fft- | 152 |
8 | Petr | 146 |
9 | orz | 145 |
10 | pajenegod | 144 |
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Auto comment: topic has been updated by shrohit_007 (previous revision, new revision, compare).
Here $$$\pi(n)$$$ denotes prime counting function, i.e number of primes not greater than $$$n$$$.
Numbers form $$$p^3$$$ can be easily counted, it's just $$$\pi \left(\lfloor \sqrt[3]{n} \rfloor\right)$$$.
To calculate numbers form $$$pq$$$ recall, that
.
Primes and $$$\pi$$$ up to $$$\sqrt{n}$$$ can be calculated straightforward using Eratosphenes sieve.
Also you can calculate values of $$$\pi\left( \lfloor \frac{n}{k} \rfloor \right)$$$ for all $$$k \geqslant 1$$$ using $$$O(n^{2/3})$$$ time as described here https://mirror.codeforces.com/blog/entry/91632
IMHO this question is not very simple as it requires some knowledge.