hdu 1018 Big Number

Big Number

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

Total Submission(s): 17743    Accepted Submission(s): 7956


[align=left]Problem Description[/align]
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.

 

 

[align=left]Input[/align]
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.

 

 

[align=left]Output[/align]
The output contains the number of digits in the factorial of the integers appearing in the input.

 

 

[align=left]Sample Input[/align]

2
10
20

 

 

[align=left]Sample Output[/align]

7
19

 

 

 

    公式:

 N ! = (sqrt(2*PI*n)) * (n/e)^n  ;

 直接求 lg(N !)  就 OK了

  代码如下 ：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define e 2.718281828459
#define PI 3.1415926

int main()
{
int n,key;
scanf("%d",&n);
while(n--)
{
scanf("%d",&key);
double x=2*PI*key;
double y=key/e;
int z;
z=(int)(  ( (  log(x)/log(10)  ) /2 )+(key*(log(y)/log(10))));
printf("%d\n",z+1);
}
return 0;
}