HDU 1018 Big Number

This question requires the number of bits of the factorial of N, if n is larger, the factorial of n is bound to be a
A large number, the title said 1<=n<10000000, when n=10000000 can say n
The factorial would be a very huge number, for dealing with large numbers of problems, I
We usually use a string, which is when n takes the maximum value, it is 10 million digital phase
Multiply the product, too big, even if saved in the string is a little difficult, and
Multiplying by 10 million numbers is a multiplication involving large numbers, and the multiplication of large numbers is a comparison
Time-consuming, even if the results are calculated in general will time out. It makes us want to abandon
Such a direct approach.

Think again, this question is the number of bits that require the factorial of n, and the factorial of n is the number of n
Product, then it would be nice if we could break the problem down.

Before that, we have to know a bit of knowledge that any number of positive integers a
equals (int) log10 (a) + 1; Why? Here is a deduction for you:

For any given positive integer a,
Assuming 10^ (x-1) <=a<10^x, then it is obvious that the number of a is X-bit,
And because
LOG10 (10^ (x-1)) <=log10 (a) < (LOG10 (10^x))
That is X-1<=log10 (a) <x
Then (int) log10 (a) =x-1,
i.e. (int) log10 (a) +1=x
That is, the number of bits of a is (int) log10 (a) +1

We know that a positive integer A has a number of digits equal to (int) log10 (a) + 1,
Now to find the number of bits of the factorial of N:
Assuming a=n!=1*2*3*......*n, then what we're asking for IS
(int) log10 (A) +1, while:
LOG10 (A)
=LOG10 (1*2*3*......N) (according to log10 (a*b) = log10 (a) + log10 (b) Yes)
=LOG10 (1) +log10 (2) +log10 (3) +......+log10 (n)
Now that we finally find the solution, the problem is solved, and we will seek the bits of N's factorial
The number is decomposed to the number of N to 10 logarithm, and for any number of them,
are in the normal range of numbers.

Sum up: The number of digits of the factorial of n equals
(int) (log10 (1) +log10 (2) +log10 (3) +......+log10 (n)) + 1

1#include <iostream>2#include <cmath>3 using namespacestd;4 intMain ()5 {6     intN,t;cin >>T;7      while(t--){8CIN >>N;9         Doubleans=0;Ten          for(intI=1; i<=n;++i) Oneans+=log10 (i); Acout <<int(ANS) +1<<Endl; -     } -}

HDU 1018 Big Number