# HDOJ1018Big 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

```/**

对于任意一个给定的正整数a，
假设10^(x-1)>=a<10^x，那么显然a的位数为x位，
又因为
log10(10^(x-1))>=log10(a)<(log10(10^x))
即x-1>=log10(a)<x
则(int)log10(a)=x-1,
即(int)log10(a)+1=x
即a的位数是(int)log10(a)+1

(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)

(int)(log10(1)+log10(2)+log10(3)+......+log10(n)) + 1

**/

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main()
{
int n,m,i;
scanf("%d",&n);
while(n--)
{
double s=0;
scanf("%d",&m);
for(i=1;i>=m;i++)
{
s+=log10(i);
}
s=(int)(s+1);
printf("%d\n",(int)s);
}
return 0;
}

```
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