2015北京网络赛 H题 Fractal 找规律
2015-09-20 20:18
375 查看
Fractal
Time Limit: 1 SecMemory Limit: 256 MB
题目连接
http://hihocoder.com/contest/acmicpc2015beijingonline/problem/8Description
This is the logo of PKUACM 2016. More specifically, the logo is generated as follows:1. Put four points A0(0,0), B0(0,1), C0(1,1), D0(1,0) on a cartesian coordinate system.
2. Link A0B0, B0C0, C0D0, D0A0 separately, forming square A0B0C0D0.
3. Assume we have already generated square AiBiCiDi, then square Ai+1Bi+1Ci+1Di+1 is generated by linking the midpoints of AiBi, BiCi, CiDi and DiAi successively.
4. Repeat step three 1000 times.
Now the designer decides to add a vertical line x=k to this logo( 0<= k < 0.5, and for k, there will be at most 8 digits after the decimal point). He wants to know the number of common points between the new line and the original logo.
Input
In the first line there’s an integer T( T < 10,000), indicating the number of test cases.Then T lines follow, each describing a test case. Each line contains an float number k, meaning that you should calculate the number of common points between line x = k and the logo.
Output
For each test case, print a line containing one integer indicating the answer. If there are infinity common points, print -1.Sample Input
30.375
0.001
0.478
Sample Output
-1
4
20
HINT
题意给你一个正方形,然后下一个正方形是这个正方形的边上的终点连起来的,然后接着这样构造下去
然后划一条线,问你经过了多少个点
题解:
找规律题,怕爆精度,所以用的java
代码:
import java.io.*; import java.math.*; import java.util.*; public class Main { static BigDecimal a[] = new BigDecimal[1005]; public static void main(String argv[]) throws Exception { Scanner cin = new Scanner(System.in); a[0] = BigDecimal.valueOf(0); BigDecimal x = BigDecimal.valueOf(0.5); BigDecimal kk = BigDecimal.valueOf(0.5); BigDecimal y; for(int i = 1 ; i <= 1000 ; ++ i) { x = x.multiply(kk); a[i] = a[i-1].add(x); } int Case = cin.nextInt(); while(Case != 0) { y = cin.nextBigDecimal(); int L = 0 , R = 500; while(L <= R) { int mid = L + (R-L) / 2; int result = a[mid].compareTo(y); if(result == 1) R = mid - 1; else if(result == -1) L = mid + 1; else { L = mid; break; } } if(a[L].compareTo(y) == 0) System.out.println(-1); else System.out.println(L*4); Case--; } } }
相关文章推荐
- hihoCoder 1227 The Cats' Feeding Spots(暴力)——ACM-ICPC国际大学生程序设计竞赛北京赛区(2015)网络赛
- HTTP劫持后续查查xjcf168.com
- 2015北京网络赛 A题 The Cats' Feeding Spots 暴力
- Http Cookies 和相关概念
- TCP可靠传输的实现
- https://leetcode.com
- 2015 网络赛
- hihoCoder 1234 Fractal——ACM-ICPC国际大学生程序设计竞赛北京赛区(2015)网络赛
- 转 TCP中的序号和确认号
- 计算机网络之面试常考
- 计算机网络之局域网&以太网
- httpd2.4特性
- Android中的Http通信
- 当打开sql server2008打开,出现报错“评估期已过。有关如何升级的测试版软件的信息请访问,http://www.....”
- 2015北京赛区网络赛模拟题
- flex http请求error#1090XML 分析器失败: 元素格式不正确。
- 2015年ACM北京网络赛 B题 (模拟+枚举题意请原谅我英语不好。。。)
- 程序一启动检查网络,如果没有网络就退出程序
- ACM-ICPC国际大学生程序设计竞赛北京赛区(2015)网络赛A题2015-9-20
- python keras (一个超好用的神经网络框架)的使用以及实例