UVa 10916 Factstone Benchmark (数学&阶乘的处理技巧)

10916 - Factstone Benchmark

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=99&page=show_problem&problem=1857

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 integern 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

Output for Sample Input

3
8

思路：字节数 k = （year - 1940) / 10,  问题就转化成 n ! < 2 ^ k < (n + 1) !, 如果单纯模拟会溢出， 所以我们对两边同取对数，因为log(a*b) = log(a) + log(b)，所以log(n!) = sum(log(i)), ( 1<= i <= n), 只要找到最小的sum(log(i)) > k * log（2） ，答案就是i- 1.

完整代码：

/*0.286s*/

#include<cstdio>
#include<cmath>
const double log_2=log(2.0);

int main(void)
{
	int year;
	while (scanf("%d", &year), year)
	{
		int n = (year - 1940) / 10;
		double k = pow(2, n) * log_2, sum = 0;
		for (int i = 1;; i++)
		{
			sum += log(i);
			if (sum > k)
			{
				printf("%d\n", i - 1);
				break;
			}
		}
	}
	return 0;
}

打表：

/*0.012s*/

#include<cstdio>

const int ans[21] =
{
	3, 5, 8, 12, 20,
	34, 57, 98, 170,
	300, 536, 966, 1754,
	3210, 5910, 10944, 20366,
	38064, 71421, 134480, 254016
};

int main(void)
{
	int year;
	while (scanf("%d", &year), year)
		printf("%d\n", ans[(year - 1960) / 10]);
}