Factstone Benchmark
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 5577   Accepted: 2524

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?

Input

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.

Output

For each test case, output a line giving the Factstone rating.

Sample Input

1960
1981
0

Sample Output

3
8

Source

OJ-ID:
poj-2661

author:
Caution_X

date of submission:
20191010

tags:
math

description modelling:
给定一个数x(x可以达到256位),找到一个数n满足n!<=x&&(n+1)!>x

major steps to solve it:
1.因为x可以达到256位,所以我们同时对n!<=x和(n+1)!>x去对数
2.得到log(n)+log(n-1)+.......+log(1)<=log(x)&&log(n+1)+log(n)+......+log(1)>log(x)

AC Code:

#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
int n;
double w;
while(~scanf("%d",&n)&&n) {
w=log();
for(int i=;i<=n;i+=) {
w*=;
}
int i=;
double f=;
while(f<w) {
f+=log((double)++i);
}
printf("%d\n",i-);
}
}

最新文章

  1. [PHP源码阅读]empty和isset函数
  2. Linux运行等级,根目录,文件类型etc
  3. iOS逆传值的三种方式
  4. BI如何让企业管理从信息化迈向智能化 ——暨珠海CIO协会成立大会圆满召开
  5. ActiveMQ使用总结
  6. C蛮的全栈之路-node篇(二) 实战一:自动发博客
  7. 如何修改WAMP中mysql默认空密码&amp;重新登录phpmyadmin
  8. jstack(查看线程)、jmap(查看内存)和jstat(性能分析)命令
  9. 360的IM可能会是什么样?
  10. 利用fiddler将本地网页放到某个域下
  11. python基础(四)字符串处理
  12. Python 一个奇特的引用设定
  13. java学习——递归
  14. shell脚本之获取CPU使用率
  15. cd 命令
  16. C# 使用Xamarin开发Android应用程序
  17. TimeSpan格式化字符串格式(摘)
  18. git Push failed: Could not read from remote repository 解决方案
  19. 牛客国庆集训派对Day1 B. Attack on Titan
  20. Codeforces Round #239 (Div. 1) 二项式差分

热门文章

  1. 使用pytorch时所遇到的问题总结
  2. Django之Django快速体验
  3. MySQL的5大引擎及优劣之分
  4. 数据库系统原理(第四章:SQL与关系数据库基本操作 )
  5. centos7 apache后台转nginx后台
  6. Android native进程间通信实例-binder篇之——解决实际问题inputreader内建类清楚缓存
  7. bayaim_java_入门到精通_听课笔记bayaim_20181120
  8. 连续线性空间排序 起泡排序(bubble sort),归并排序(merge sort)
  9. PHP转Go系列:字符串
  10. CodeForces - 1250L(思维+贪心)