hdu 1018 Big Number(纯数学公式)

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17227    Accepted Submission(s): 7705

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

Source

Asia 2002, Dhaka (Bengal)

Recommend

JGShining

 

思路：///N!的位数=log(N!)/log(10);log(N!)=N*log(N)-N+0.5*log(2*m*pi);

 

失误点：注意函数重载，别CE了

 

///N!的位数=log(N!)/log(10);log(N!)=N*log(N)-N+0.5*log(2*m*pi);
#include<iostream>
#include<cmath>
using namespace std;
const double pi=3.1415926;
double sum;
int lowbit(int m)
{
  double s=m*log((double)m)-m+0.5*log((double)2*m*pi);
  s=s/log((float)10);
  return (int)s+1;///如果是7.5位，则说明实际上是8位
}
int main()
{
  int cas;
  while(cin>>cas)
  {
    while(cas--)
    { int m;
      cin>>m;
      cout<<lowbit(m)<<"\n";
    }
  }
}