hdu--1018--Big Number（斯特林公式）

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37482    Accepted Submission(s): 18009

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

 
    Name: hdu--1018--Big N 
    Copyright: ©2017 日天大帝 
    Author: 日天大帝 
    Date: 20/04/17 21:03
    Description: 两种思路：1，老老实实取对数     
　　　　　　　　　　　　　　　2，斯特林公式（Stirling's approximation）是一条用来取n的阶乘的近似值的数学公式。一般来说，当n很大的时候，n阶乘的计算量十分大，所以斯特林公式十分好用，而且，即使在n很小的时候，斯特林公式的取值已经十分准确。
*/
#include<iostream>
#include<cmath>
using namespace std;
int main(){
    int n;cin>>n;
    while(n--){
        int m;cin>>m;
        double ct = 0;//log10()取对数要用double存 
        for(int i=1; i<=m; ++i){
            ct += log10((double)i);
        }
        cout<<(int)ct+1<<endl;
    }
    return 0;
}