POJ 1423:big Number Math

Big number

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 27537		Accepted: 8787

Description in many applications very large integers numbers is required. Some of these applications is using keys for secure transmission of data, encryption, etc. In this problem you is given a number, you has to determine the number of digits in the factorial of the number.

Input input consists of several lines of the integer numbers. The first line contains a integer n, which is the number of cases to being tested, followed by n lines, one integer 1 <= M <= 10^7 on each line.

Output the output contains the number of digits in the factorial of the integers appearing in the input.

Sample Input

2
20

Sample Output

7
19

Ideas:

Seek log10 (n!) for the number of digits of the factorial of N. =LOG10 (1) +log10 (2) +......log10 (n)

another formula is given in the art of computer programming.

n! = sqrt (2*π*n) * ((n/e) ^n) * (1 + 1/(12*n) + 1/(288*n*n) + O (1/n^3))

π= ACOs ( -1)

e = exp (1)

10 logarithm on both sides

Ignore log10 (1 + 1/(12*n) + 1/(288*n*n) + O (1/n^3)) ≈log10 (1) = 0

Get the formula

log10 (n!) = log10 (sqrt (2 * pi * n)) + N * log10 (n/e) . Baidu to the ... Forgive me for not being good at maths ~ ~ ~ Ac-code:

#include <iostream>
#include <cmath>
using namespace std;
Double E=exp (1.0), Pi=acos ( -1.0);
int main ()
{
	int T;
	Long long N,ans;
	cin>>t;
	while (t--)
	{
		cin>>n;
		Ans=int (log10 (sqrt (2.0*pi*n)) +n*log10 (n/e)) +1;
		cout<<ans<<endl;
	 } 
	 return 0;