poj 2661 Factstone Benchmark (Stirling数)

//题意是对于给定的x，求满足n! <= 2^(2^x)的最大的n
//两边同取以二为底的对数，可得： lg2(n!) <= 2^x

 1.
   log2(n!) = log2(1) + log2(2) + .. + log2(n);一个循环即可

 2.
   Stirling

    

#include <iostream>

#include <string>

#include<sstream>

#include <cmath>

#include <map>

#include <stdio.h>

#include <string.h>

#include <algorithm>

using namespace std;

// #define  LL __int64

#define  LL long long

const double pi=acos(-1.0);

const double inx_2=log(2.0); // log -> inx  log10

int f[30];

int main()

{

  f[0]=1;

  int i;

  for(i=1;i<30;i++)

    f[i]=f[i-1]*2;

  int year;

  while(scanf("%d",&year),year)

  {

      double t=f[(year-1960)/10+2];

      double s=(0.5*log(2*pi)+log(1.0)-1)/inx_2;

      i=1;

      while(s<=t)

      {

         i++;

         s=(0.5*log(2*pi*i)+i*log((double)i)-i)/inx_2;

      }

      printf("%d\n",i-1);

  }

  return 0;

}