ACM HOJ 1018 Big Number

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10639    Accepted Submission(s): 4780

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

 

Source

Asia 2002, Dhaka (Bengal)

 

Recommend

JGShining

斯特灵公式是一条用来取n阶乘近似值的数学公式。一般来说，当n很大的时候，n阶乘的计算量十分大，所以斯特灵公式十分好用，而且，即使在

 

n很小的时候，斯特灵公式的取值已经十分准确。

公式为：

这就是说，对于足够大的整数n，这两个数互为近似值。更加精确地：

或者：

 

 

 

超强大的公式

n！=（2*pi*n)^1/2 *(n/e)^n *e^(a/12*n)

#include<iostream>

#include<cmath>

using namespace std;

const double PI=acos(-1.0);

const double e=exp(1.0);

int main()

{

    int T,n;

    cin>>T;

    while(T--)

    {

        cin>>n;

        if(n!=1)

          cout<<(int)(0.5*log10(2*PI*n)+n*log10(n/e)+1)<<endl;

        else cout<<1<<endl;

    }

    return 0;   

}   

 

 

#include<stdio.h>
#include<math.h>
using namespace std;
int main()
{
    int n,i;
    double sum;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        sum=1;
        scanf("%d",&n);
        if(n==1){printf("1\n");continue;}
        for(i=2;i<=n;i++)
        {
            sum+=log10((double)i);
        }    
        printf("%d\n",(int)sum);
    }
    return 0;    
}    

#include<cmath>
#include<stdio.h>
#include<iostream>
using namespace std;
const double  PI=acos(-1.0);
const double e=exp(1.0);
int main()
{
    int T,n,i;
    int cnt;
    scanf("%d",&T);
    while(T--)
    {
            scanf("%d",&n);
            if(n==1){printf("1\n");continue;}//这个不能少，否则会错误，WR了好几次 
            cnt=(int)(0.5*log10(2*PI*n)+n*log10(n/e)+1);
            printf("%d\n",cnt);
        
    }    
    return 0;
}    

#include<stdio.h>
#include<cmath>
using namespace std;
const double PI=acos(-1.0);
const double e=exp(1.0);

int main()
{
    int n,cnt;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        cnt=(int)(0.5*log10(2*PI*n)+n*log10(n/e))+1;
        printf("%d\n",cnt);
    }
    return 0;    
}