【水题-数学】 HDU 1141 Factstone Benchmark

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

8

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;}