poj2661(Stirling公式)

Factstone Benchmark

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4699		Accepted: 2240

Description

Amtel has announced that it will release a 128-bit computer chip by 2010, a 256-bit computer by 2020, and so on, continuing its strategy of doubling the word-size every ten years. (Amtel released a 64-bit computer in 2000, a 32-bit computer in 1990, a 16-bit
computer in 1980, an 8-bit computer in 1970, and a 4-bit computer, its first, in 1960.) 

Amtel will use a new benchmark - the Factstone - to advertise the vastly improved capacity of its new chips. The Factstone rating is defined to be the largest integer n such that n! can be represented as an unsigned integer in a computer word. 

Given a year 1960 <= y <= 2160, what will be the Factstone rating of Amtel's most recently released chip?
Input

There are several test cases. For each test case, there is one line of input containing y. A line containing 0 follows the last test case.
Output

For each test case, output a line giving the Factstone rating.
Sample Input

1960
1981
0

Sample Output

3
8

/*
思路：通过换算找到y年的位数，即为pow(2,N)，这样能表达的最大数即为2^pow(2,N)，
化成对数形式为pow(2,N)*log(2.0)
此外，利用Stirling公式n!≈sqrt(2*pi*n)*[(n/e)^n]
化成对数形式lnN!=NlnN－N+0.5ln(2N*pi)
计算出每个数N的对应N!的对数大小
预先计算出2016年的那个对数形式大小为多少，保证最大的N对应N!对数大小大于那个数就好了
*/
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
using namespace std;
#define pi 3.1415926
#define maxn 270005
double lnN[maxn];
void init()
{
lnN[1]=0;
for(int i=2;i<maxn;i++)
{
lnN[i]=i*log(1.0*i)-i+0.5*log(2.0*i*pi);
}
}
int main()
{
init();
int y;
double N,lnN1;
while(~scanf("%d",&y)&&y)
{
N=(double)((y-1940)/10);
lnN1=pow(2,N)*log(2.0);
for(int i=2;i<maxn;i++)
{
if(lnN[i]>lnN1)
{
printf("%d\n",i-1);
break;
}
}
}
}