UVA 11762

Правка en1, от Upgrading....., 2019-02-05 03:52:11

UVA 11762

Suppose we want to calculate E(N) where N is a number,k is the number of it’s distinct prime factors and p is the number of primes < n Normal rule, E(N)=(1/p)*(E(N/a1) + E(N/a2) + E(N/a3) … + E(N/ak)) + ((p-k)/p)*(1 + E(N))

I don't understand the editorial part of this problem.... Below part is the editorial part..

E(N)=1+(1/p)*(E(N/a1) + E(N/a2) + E(N/a3) … + E(N/ak)) + ((p-k)/p) * E(N)

-Why here add 1 in the first? Why not add 1 in the failure part?

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  Rev. Язык Кто Когда Δ Комментарий
en4 Английский Upgrading..... 2019-02-05 11:10:29 2 Tiny change: 'imes < n\nNormal ' -> 'imes < n\n\nNormal '
en3 Английский Upgrading..... 2019-02-05 11:10:12 225
en2 Английский Upgrading..... 2019-02-05 11:03:58 4 Tiny change: 'N)=(1/p)*(E(N/a1) + ' -> 'N)=(1/p)*(1 + E(N/a1) + '
en1 Английский Upgrading..... 2019-02-05 03:52:11 594 Initial revision (published)