[UVA](https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2862) 11762↵
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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)*(1 + E(N/a1) + E(N/a2) + E(N/a3) … + E(N/ak)) + ((p-k)/p)*(1 + E(N))↵
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I don't understand the editorial part of this problem....↵
Below part is the editorial part..↵
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E(N)=1+(1/p)*(E(N/a1) + E(N/a2) + E(N/a3) … + E(N/ak)) + ((p-k)/p) * E(N)↵
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-Why here add 1 in the first? Why not add 1 in the failure part?
↵
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)*(1 + 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?
