【水题-数学】 HDU 1141 Factstone Benchmark

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本文介绍了一种计算Factstone Benchmark值的方法，通过求解n! < 2^(2^p)来确定最大整数n，其中p=(年份-1960)/10+2。利用对数简化计算过程，适用于从1960到2160年的年份。

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1141 ( Factstone Benchmark )

Problem 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?
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. For each test case, output a line giving the Factstone rating.

Sample Input
1960
1981
0

Sample Output
3

说白了，这题就是让你求最大的n，使其满足n!<2^(2^p)，这里的p需要先计算出(p=(年份-1960)/10+2)。但是很明显，当年份为2160时，p=22，那么2^(2^p)是多少呢？这个算起来十分麻烦，数据超过了double的范围。那要怎么办呢？能不能把数字变小一些。仔细观察2^(2^p)，有两次指数，能不能去掉一次呢？当然可以，取对数！等式两边取对2的对数，就变成了log2(1)+log2(2)+log2(3)+...+log2(n)<2^p。这样就很容易计算了。

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
    int n;
    while (cin >> n, n)
    {
        int p = (n - 1960) / 10 + 2;
        p = 1 << p;
        double sum = 0.0, fac = 1.0;
        while (sum < (double)p)
            sum += log(fac) / log(2.0), fac += 1.0;
        cout << int(fac - 2.0) << endl;
    }
    return 0;
}