Factstone Benchmark
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1951 Accepted Submission(s): 1078
Total Submission(s): 1951 Accepted Submission(s): 1078
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.
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
Source
题意:计算机芯片的bit每年都不一样,求一个数n使得n!<=2^bit时n的最大值。
思路:换成log计算log(n!) = log(n) + log(n-1) + ...... + log(1);
#include <stdio.h>
#include <math.h>
int main()
{
int n, bit, i;
while(scanf("%d", &n) != EOF){
if(!n) break;
bit = (n - 1960) / 10 + 2;
bit = 1<<bit;
double b = 1.0 * bit * log(2.0), s = 0;
for(i = 1;;i++){
s += log((double)i);
if(s > b){
printf("%d\n", i - 1);
break;
}
}
}
}