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

时间:2023-03-08 16:24:31

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\n", i - 1);
break;
}
}
}
return 0;}