ACM--steps--2.3.3--stirling公式求n！

Big Number

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 955 Accepted Submission(s): 643

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]

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

[align=left]Sample Input[/align]

2
10
20

[align=left]Sample Output[/align]

7
19

[align=left]Source[/align]
Asia 2002, Dhaka (Bengal)

[align=left]Recommend[/align]
JGShining

第一个代码，就是求n！算出阶乘的位数，要求出位数，取完对数后最后要+1；

第二个代码，运用stirling公式求解n！；

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int n,a;
cin>>n;
while(n--)
{
cin>>a;
double sum=0;
for(int i=1;i<=a;i++)
{
sum+=log10(i);
}
cout<<(int)(sum)+1<<endl;
}
return 0;
}

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
double PI=acos(double(-1));//获取PI的值，对-1取反余弦函数。
double e=exp(double(1));//获取自然底数e的值
//运用stirling公式，求取n！的值。
int T,a;
cin>>T;
double ans;
while(T--)
{
ans=0;
cin>>a;
ans+=0.5*log10(2.0*a*PI)+a*(log10((double)a)-log10(e));
cout<<(int)ans+1<<endl;
}
return 0;
}