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[];
int main()
{
  f[]=;
  int i;
  for(i=;i<;i++)
    f[i]=f[i-]*;
  int year;
  while(scanf("%d",&year),year)
  {
      double t=f[(year-)/+];
      double s=(0.5*log(*pi)+log(1.0)-)/inx_2;
      i=;
      while(s<=t)
      {
         i++;
         s=(0.5*log(*pi*i)+i*log((double)i)-i)/inx_2;
      }
      printf("%d\n",i-);
  }
  return ;
}