poj1423---Ask for a large number of digits, I guess the algorithm that counts less than how many bits of input characters on the site

Law one: The logarithm of a number, +1 rounding is its number of digits

Problem converted to int (log10 (n!) +1), logarithmic property log10 (n!) =LOG10 (N) +log10 (N-1) +...+log10 (1)

/* Use LOG10 to find the number of digits */#include <stdio.h> #include <math.h>int main () {    int tim,n;    scanf ("%d", &tim);    while (tim--)    {        int i;        Double numofdigit=1;        scanf ("%d", &n);        for (i=n;i>=1;i--)        {            numofdigit+=log10 (i);        }        printf ("%d\n", (int) numofdigit);}    }

When n is large, time is long, TLE

Law II: Stirling formula

Log (n!) = log10 (sqrt (2*pi*n)) + N*LOG10 (n/e);

/* Use Stirling to find the number of positions */#include <stdio.h> #include <math.h> #define e 2.718281828459045#define pi 3.141592653589793239int Main () {    int t;    scanf ("%d", &t);    while (t--)    {        int n;        scanf ("%d", &n);        Double num_digit;        Num_digit=log10 (sqrt (2*pi*n)) + N*LOG10 (n/e);        printf ("%d\n", (int) num_digit+1);    }    return 0;}

WA two times reason:

Num_digit=log10 (sqrt (2*pi*n)) + n*log10 (n/e) +1;        printf ("%d\n", (int) num_digit);

When N=1,num_digit=0, because when N=1,

log10 (sqrt (2*pi*n)) + n*log10 (n/e) =-0.03

The last value is 0.

Summary: When using the Stirling formula to calculate the number of digits, taking into account the n=1, plus 1 placed after taking the whole

(int) (Log10 (sqrt (2*pi*n)) + N*LOG10 (n/e)) =0
Plus 1, after taking the whole place.

Int (3.1+1) =4
Int (3.1) +1=4
Int (3) +1=4
Int (3+1) =4
Int (-0.03) +1=1

poj1423---Ask for a large number of digits, I guess the algorithm that counts less than how many bits of input characters on the site