Computes the number of digits of the factorial of a large number n (decimal) (log I accumulation method)

Input:

Enter 1 positive integers per line n, (0<n<1000 000)

Output:

For each n, the number of (decimal) digits of the output n!.

Analysis:

This problem uses brute force method. By definition, direct solution!

The so-called n! decimal digits, is log (n) +1, according to the mathematical formula is: n!=1*2*3*.....*n;

LG (n!) =LG (2) +......LG (n);

Code:

Enter a number n, and you calculate the number of decimal digits of the factorial of the digit width//For example: 3! =6, the width is 1//sample data://n=3  output 1//n=32000  output 130271//n=1000000  output 5565709#include <string> #include < iostream> #include <iomanip> #include <stdio.h> #include <cmath>using namespace Std;int main () {    long int n;    long int i;    Double sum;    while (scanf ("%ld", &n)!=eof)    {        sum=0.0;        for (i=2; i<=n; i++)        {            sum+=log10 (i);        }        printf ("%ld\n", (int) sum+1);    }    return 0;}

Computes the number of digits of the factorial of a large number n (decimal) (log I accumulation method)