HDU 1141

Factstone Benchmark

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2259    Accepted Submission(s): 1277

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.

Sample Input

1960

1981

0

Sample Output

3

8

题意：××公司是制造computer的，1960年它造的computer是4bit的，之后每10年翻倍；

有一个衡量computer的标准，就是它最大可以存下n!(无符号位)，那么它就是n级；

求x年时该公司的电脑为几级（1960<= x <=2160)；

思路：也是求位数的题目，方法类似hdu1018 链接http://www.cnblogs.com/ISGuXing/p/7301041.html

只不过用log2

这道题打表。

代码：

#include<iostream>
#include<stdio.h>
#include<cmath>
#include<algorithm>
#include<math.h>
using namespace std;
int a[350000];
int main(){
    double k=0.0;
    for(int i=1;i<=350000;i++){
        k+=log2(i);
        a[i]=(int)k+1;
    }
    int year;
    while(scanf ("%d",&year)&&year!=0){
        int x=(year-1960)/10;
        int bit=4*(pow(2,x));
        int pos=lower_bound(a,a+350000,bit)-a;
        if(a[pos]>bit){
            pos--;
        }
        cout<<pos<<endl;
    }
    return 0;
}