hdu1018 Big Number (阶乘的位数)

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14952    Accepted Submission(s): 6631

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 ≤ 10⁷ 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：给你一个 n ，计算 n！ 有多少位数

分析：

    123456=1.23456*10^5;
    log10(123456)=5.09151;
    log10(1.23456*10^5)=log10(1.23456)+log10(10^5)=0.09151+5;
    故int(log10(n))+1 就是n的位数

 

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    int n,test,i,ans;
    double t;
    cin>>test;
    while(test--)
    {
        cin>>n;
        t=0;
        for(i=2;i<=n;i++)
            t+=log10(i*1.0);
        ans=int(t)+1;
        cout<<ans<<endl;
    }
    return 0;
}