hdu1018（求n！的位数）

Posted 2022-05-27 martinue

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 ≤ 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
  

题意读懂了之后就是求位数的题目嘛，求一个数n的位数很容易联想到（int）log10（n）+1，现在变成了n的阶乘，恰好又log里面相乘，直接变相加呀！这种水题不造为毛会在亚洲赛出现。。。。

貌似还有一种方法是套公式，斯特林公式：lnN!=NlnN-N+0.5*ln(2*N*pi)。

#include <iostream>
#include <stdio.h>
#include<string.h>
#include<math.h>
using namespace std;
int main()

    int n,t;scanf("%d",&t);
    while(t--)
    
        scanf("%d",&n);
        double s=0;
        for(int i=1;i<=n;i++)
            s+=log10(i);
        printf("%d\\n",(int)s+1);
    
    return 0;