poj 1423 Big Number

Big Number http://poj.org/problem?id=1423
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 24637 Accepted: 7895

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 <= m <= 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

Source

Dhaka 2002 思路：log10(n!)=log10(1*2*3…*n)=log10(1)+log10(2)+…+log10(n)

方法一：用空間換時間，開個數組把結果存起來一部分！

#include"stdio.h"
#include"math.h"
#include"string.h"
#define N 8000000
double a[N];
void Inti()
{
	int i;
	a[0]=0;
	for(i=1;i









方法二：
n! = sqrt(2*π*n) * ((n/e)^n) * (1 + 1/(12*n) + 1/(288*n*n) + O(1/n^3)) π = acos(-1) e = exp(1) 兩邊對10取對數 

忽略log10(1 + 1/(12*n) + 1/(288*n*n) + O(1/n^3)) ≈ log10(1) = 0 得到公式 

log10(n!) = log10(sqrt(2 * pi * n)) + n * log10(n / e)。







#include"stdio.h"

#include"math.h"

#include"string.h"

#define N 80000

int main()

{

int t,n,i;

double s,pi,e;

pi=acos(-1);

e=exp(1);

scanf("%d",&t);

while(t--)

{

scanf("%d",&n);

s=log10(sqrt(2*pi*n))+n*log10(n/e);

if(n==1)

s=1;

printf("%d\n",(int)ceil(s));

}

return 0;

}