hdoj 1018 Big number 求n阶乘的位数

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24102    Accepted Submission(s): 10939

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:      求n阶乘的位数    （n可以是很大的数）

思路：    1、将很多大整数的乘法问题转换为加法。         使用取对数的方法。            

                 2、某数取以十为底的对数，所得的数的整数位+1就是这个数的位数。

代码：

#include <iostream>
#include <math.h>


using namespace std;


int main()
{
    long long t,n;
    double i;
    long long m;
    double s;
    cin>>t;
    while(t--)
    {
        s=0;
        cin>>n;
        for(i=1.0;i<=n;i++)
        {
            s+=log10(i);
        }
        m=(long long)s;
        cout<<m+1<<endl;
    }
    return 0;
}