Fast computing n! The number of prime factors

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Learning
us a sample to explain:

12345678 8 we find! 2 the number
　111 First we calculate the number of multiples of 2: 8/2 = 4
　　　. 1. 1 Next we calculate the number of a multiple of 4: 8/4 = 2
　　　　　　1 Last we solve a third the number of layers is 2: 8/8 = 1

We have 4 + 2 + 1 = 7, so there have been a total of seven 2.

which is:
$cnt (x) = [n / (x ^ 1)] + [n / (x ^ 2)] + [n / (x ^ 3)] + ...$ (Until power x is larger than n)
Here we can see: The method we usually find that a one demand (that is, every number count again), and this way we each row seek each row, although the effect the same, but for up fast. worth learning.