uva 10916 Factstone Benchmark（对数函数的活用）

Factstone Benchmark

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

Output for Sample Input

3
8

题目大意：给出年份，每个10年对应一个当前计算机可支持的字节位数，计算n! < max（max 为当前计算机能表示的最大整数），求最大n.

解题思路：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.

#include<stdio.h>
#include<math.h>

int main(){
	int year;
	while (scanf("%d", &year), year){
		int n = (year - 1940) / 10;
		double k = pow ( 2, n) * log10(2), sum = 0;
		for (int i = 1; ; i++){
			sum += log10(i);
			if (sum > k){
				printf("%d
", i - 1);
				break;
			}
		}
	}
	return 0;}