Calculates the number of digits of the factorial of a large number n (decimal) reproduced

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:

1234567891011121314151617181920212223242526272829303132

//输入一个数字n，请你计算该数的阶乘的十进制数的位数宽度//比如：3！=6， 则宽度为1//样例数据：//n=3  输出1//n=32000  输出130271//n=1000000  输出5565709#include <string>#include <iostream>#include <iomanip>#include <stdio.h>#include <cmath>usingnamespacestd;intmain(){    longintn;    longinti;    doublesum;    while(scanf("%ld", &n)!=EOF)    {        sum=0.0;        for(i=2; i<=n; i++)        {            sum+=log10(i);        }        printf("%ld\n", (int)sum+1 );    }    return0;}

Category: Number theory, mind + logic + thinking

Calculates the number of digits of the factorial of a large number n (decimal) reproduced