POJ 1423 Big Number 大数阶乘的位数计算

Big Number

Time Limit:1000MS		Memory Limit:65536K
Total Submissions:17518		Accepted:5541

Description

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

Input

Input consists of several lines of integer numbers. The first line contains an integer n, which is the number of cases to be 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
10
20

Sample Output

7
19

我们知道整数n的位数的计算方法为：log10(n)+1

故n!的位数为log10(n!)+1

由于这是高精度没法用普通方式计算阶乘，否则会TLE

可以用斯特林(Stirling)公式求解

斯特林（Stirling）公式：

于是求n!的位数就是求log10((2*PI*n)^1/2*(n/e)^n)+1

即  1/2*log10(2*PI*n)+n*log10(n/e)+1

/* Author yan
*/
#include<stdio.h>
#define PI 3.141592654
#define E 2.71828182846
int l(int n)
{
	int s=1;
	if(n>3)
		s=log10(2*PI*n)/2+n*log10(n/E)+1;
	return s;
}
int main()
{
	int n;
	int _i,cache;
	scanf("%d",&n);
	for(_i=0;_i<n;_i++)
	{
		scanf("%d",&cache);
		printf("%d/n",l(cache));
	}
	return 0;
}