(eden)性能评估

时间：2015-12-16 10:50:52 阅读：250 评论：0 收藏：0 [点我收藏+]

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题目描述：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.

输入：Given a year 1960 ≤ y ≤ 2160, what will be the Factstone rating of Amtel‘s most recently released chip?

输出：For test, there is one line of input containing y. For each test case, output a line giving the Factstone rating.

示例：

输入： 1960

输出： 3

（解释：因为1960年计算机是支持4位，其中的最大的无符号数为（全为1）：15；3的阶乘为6，4的阶乘为24，故这里的n=3）

Hint:

1.取对数避免大数计算，log(n*m) = log(n) + log(m)。

2.准确理解题意。

思路

y = y / 10 - 194;

求得2^y这个指数，得到该年应该得到多少位的机，然后2^(2^y) - 1即该年的最大无符号整数，但由于直接这样计算得到的数为inf无穷大，即计算机装不下，所以要经过处理，本来是得到2^(2^y) - 1即该年的最大无符号整数这个之后，判断它在哪两个阶乘范围之内n!~(n+1)!，然而这样计算又还要写阶乘函数，又有判断循环，耗时，所以采用

p! < 2^2^y
两边取对数
log(1)+log(2)+...+log(p) < (2^y)*log(2)
即采用whlie循环一直加直到加到某一个p0超过了后面则出循环，p0 - 1即为所求p

因为如果直接用(log(2^(2^y) - 1))会超过int float double类型的范围，所以只能够这样将减一忽略不计，因为并不会影响，从而得到只需要计算2^y即可；

代码

 1 #include<stdio.h>
 2 #include<math.h>
 3 int main() {
 4     int y, i;
 5     scanf("%d", &y);
 6     y = y / 10 - 194;
 7     double k = pow(2, y);
 8     i = 1;
 9     double tot = log(1);
10     while (tot < k * log(2)) {
11         i++;
12         tot = tot + log(i);
13     }
14     printf("%d\n", i - 1);
15 }

标答

 1 // from younglee.
 2 #include<stdio.h>
 3 #include<math.h>
 4 int main(void) {
 5     int year, num, n = 2;
 6     double sum = 0;
 7     // for input
 8     scanf("%d", &year);
 9  
10     // for processing.
11     num = pow(2, (year - 1960)/10 + 2);//也是求Index
12     while (sum <= num) {
13         sum += log(n) / log(2);//把log(2)移到另一边
14         ++n;
15     }
16  
17     // for output.
18     printf("%d\n", n - 2);//因为最后还加了1
19     return 0;
20 }