HDU1018-Big Number

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 36450    Accepted Submission(s): 17456

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

 

思路：

一个数的位数等于他对10取对数+1

源代码：

#include<cstdio>
#include<cmath>
#include<iostream>
using namespace std;
int main()
{
    int t;
    int num;
    double sum;
    scanf("%d",&t);
    while(t--)
    {
        sum=0;
        scanf("%d",&num);
        for(int i=1;i<=num;i++)
        {
            sum+=log10((double)i);
        }
        printf("%d\n",(int)sum+1);
    }
    return 0;
}