2661 Factstone Benchmark 求m! < 2^n， n是已知, 求满足该式最大的m

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本文介绍了一种新的计算机性能评测标准——Factstone基准，并详细解释了如何根据年份计算出Amtel公司发布的不同位数计算机芯片所能达到的最大Factstone评分。

Factstone Benchmark

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2323		Accepted: 1251

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

// 求m! < 2^n， n是已知, 求满足该式最大的m
// 两边取对数就成了log2(1)+log2(2)+...log2(m) < n， 其中log2(m)=logm/log2
// 后来突然想起, 原来曾经囧在这题上.
#include <iostream>
#include <cmath>
#include <cstdio>
using namespace std;

int main ()
{
 int y, n, m;
 while (scanf("%d", &y) ==1 && y)
 {
  n = 2;
  while (y >= 1960) y -= 10, n *= 2;
  double sum = 0;
  for (m = 1; sum + log((double)m) / log(2.0) < n; m++)
  {
   sum += log((double)m) / log(2.0);
  }
  printf("%d/n", m - 1);
 }
}