题目大意： 给出一个年份，通过和1960进行计算求出第y年的word数。然后求出
2^word>n! 的最大的n
解法： 因为n！比较大 所以两边先取2的对数
word>log(2)1+log(2)2+....log(2)n
再利用换底公式 log(2)n=logn/log2 进行累加
Problem B: 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
1 #include <cstdio> 2 #include <cmath> 3 int main() 4 { 5 int year; 6 while(scanf("%d",&year)==1) 7 { 8 if(year==0)break; 9 year-=1960; 10 year = (year-year%10)/10; 11 int word=1; 12 for(int i=1;i<=year;i++) 13 word*=2; 14 15 word*=4; 16 17 int ans=1; 18 double sum=0; 19 while(1) 20 { 21 sum+=log(ans)/log(2); 22 if(word<=sum)break; 23 else 24 ans++; 25 } 26 printf("%d ",ans-1); 27 28 } 29 30 31 return 0; 32 }