POJ 1423 Big Number

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 <= m <= 10^7 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

（部分转载）

补充知识 ：若n=a+b^10,(a<1,b∈N)，则n的位数为b。

所以要求一个数字num有多少位，可以用(int)log10(num)+1，这样就求出num有多少位.

如果sum=n*(n-1)(n-2)……*1，那么log10(sum)=log10(n)+

log10(n-1)+…..+log10(2)+log10(1)。即sum的位数是

( int )( log10(n)+log10(n-1)+…..+log10(2)+log10(1) )+1。

（从小到大算）

下面这一种方法是程序外打表法，需要消耗时间。

还有一种是程序外打表，就是直接给复制，再导入，则肯定是0ms。

#include <stdio.h>
#include <math.h>
int a[10000010];
int main()
{
    double d;
    int i;
    int t;
    int n;
    for(i=1,d=0;i<10000005;i++){
        d+=log10(i);
        a[i]=(int)d+1;
    }
    scanf("%d",&t);
    while(t-->0){
        scanf("%d",&n);
        printf("%d\n",a[n]);
    }
    return 0;
}

还有一种方法：

利用Stirling公式. n!≈sqrt(2*pi*n)*[(n/e)^n]

n越大，答案越准确。

在n=1的情况下，也最大不过相差0.2.

这里只要计算位数，所以是可以用的。

此处的：

e ≈ 2.7182818284590452354, pi ≈ 3.141592653589793239;

#include<iostream>
#include<cmath>
using namespace std;
const double e = 2.7182818284590452354, pi = 3.141592653589793239;
double strling_digits_num(int n)
{
        return log10(2*pi*n)/2.0+n*(log10(n/e));
}
int main()
{
        int t;
        cin>>t;
        while(t--)
        {
                int n;
                cin>>n;
                double m=0;
                m=strling_digits_num(n);
                int answer=(int)m;
                answer++;
                cout<<answer<<endl;
        }
        return 0;
}