hdoj1018--Big Number

Problem 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 ≤ n ≤ 10⁷ 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

算法：

我们知道了一个正整数a的位数等于(int)log10(a) + 1，

现在来求n的阶乘的位数：
假设A=n!=1*2*3*......*n，那么我们要求的就是
(int)log10(A)+1，而：
log10(A)
        =log10(1*2*3*......n)  （根据log10(a*b) = log10(a) + log10(b)有）
         =log10(1)+log10(2)+log10(3)+......+log10(n)
总结一下：n的阶乘的位数等于
 (int)(log10(1)+log10(2)+log10(3)+......+log10(n)) + 1

#include<iostream>
#include<cmath>
using namespace std;

double fun(int n)
{
	return log10(n*1.0);
}

int main()
{
	int n,t,i,r;
	double s;
	while(cin>>n)
	{
		while(n--)
		{
			s=0;
			cin>>t;
			for(i=2;i<=t;i++)
			{
				s+=fun(i);
			}
			r=ceil(s);
			cout<<r<<endl;
		}
	}
	return 0;
}