hdu 1018 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

n!的位数len=log10(n!)＝log10(n)+log10(n-1)+log10(n-2)+......+log10(2)+log10(1)+1.再对len取整就可以了

直接求代码如下：差点超时

#include<stdio.h>
#include<math.h>
int main()
{
    int n,t,i;
    double count;
    scanf("%d",&t);
    while(t--)
    {
        count=0;scanf("%d",&n);
        for(i=2;i<=n;i++)
            count+=log10(i);
        printf("%d
",int(count)+1);
    }
    return 0;
}

下面介绍一个一个公式log10(n!)=1.0/2*log10(2*pi*n)+n*log10(n/e)

lnN!=NlnN－N+0.5ln(2N*pi) 

而N的阶乘的位数等于：log10(N!)取整后加1

log10(N!)=lnN!/ln(10)所以len=lnN!/ln(10)+1

#include<stdio.h>
#include<math.h>
const double pi=acos(-1.0);
int main()
{
    int t,n,count;
    scanf("%d",&t);
    while(t--)
    {
        count=0;
        scanf("%d",&n);
        count=(0.5*log(2*pi*n)+n*log(n)-n)/log(10);
        printf("%d
",count+1);
    }
    return 0;
}