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 中学生解 Google 编程赛模拟题

作者:qiushuiwuhen 时间: 2005-08-23 文档类型:原创 来自:蓝色理想

第 1 页 Google 编程赛模拟题
第 2 页 中学生一天的解答
第 3 页 本科生评价和总结

三、本科生评价和总结
第一题解法可以继续简化,如四面一起加的两次可合并
第二题解法可用二分逼近法
第三题解法利用递归是可以,但时间复杂度和空间复杂度过大,要好好的优化,如将同类操作合并:

运行代码框

[Ctrl+A 全部选择 提示:你可先修改部分代码,再按运行]

如果去掉 if(n>m)return -1; 输出结果:
BABABABABABABABABABABABAB=4.0870725092931246e+21
从21位就开始不精确,准确值=4087072509293123892361

但总的来说,完成的不错,算“良”的了

总而言之,这套题不算很难,假设一开始就想到解法(其实大部分人的坎都在这,可能花一小时,一天,甚至永远),乐观估计解题时间分别为20/10/30分(但实际上一开始不可能想的很缜密,要完善可能花三五倍的时间),但要在一小时内完成这三道题,除非胸有成竹,一次成文,看来非一等一的高手不可,也许空中网的李杨可以。如果觉得不过瘾,再加一道题


DiskDefrag(磁盘整理)

Problem Statement
????
When files are stored on a hard disk, they often become fragmented. This means that the file is not stored in sequential sectors on the disk. The first half of a file might be stored in sector 243, while the second half of a file might be far away in sector 105. The goal of defragmenting a hard drive is to arrange the files so that each file is stored in order on sequential sectors of the disk. Thus, if a file required 4 sectors of storage space, it would end up in sectors N, N+1, N+2, and N+3, for some N. Typically, this is done to increase the overall speed of the computer.  There are a number of programs that will defrag a hard disk as described above. However, many of them are painfully slow. You are trying to develop a new algorithm to defrag hard drives, but before you start, you would like to determine how fast you can defrag a very small drive without very many files on it. You will be given the locations of a number of files on a small hard disk, and are to determine the minimum number of sectors that must be moved before the entire drive is defragged. You have enough memory to hold two sectors worth of data at once, but that is all.  You will be given a String[], disk, each of whose elements represents a single file. Each element of disk will be formatted as a single-space delimited list of integers which represent the locations of the parts of the file, in order. Hence, the String, "4 9 6 59 41" represents a file stored in 5 sectors where the first part of the file is in sector 4 of the disk. One way to defrag this file would be to move the contents of sector 9 to sector 5, the contents of sector 59 to sector 7, and the contents of sector 41 to sector 8. By doing this, the file would be stored sequentially in sectors 4-8. You will also be given an int, size, representing the total number of sectors on the disk (sectors 0 through size-1, inclusive, may contain data). You are to return the smallest number of sectors that must be moved to defragment the whole disk. Keep in mind that you can not move data to a sector until any data being stored there is moved.
Definition
????
Class:
DiskDefrag
Method:
minMoves
Parameters:
String[], int
Returns:
int
Method signature:
int minMoves(String[] disk, int size)
(be sure your method is public)
????

Constraints
-
size will be between 10 and 100, inclusive.
-
disk will contain between 1 and 12 elements, inclusive.
-
Each element of disk will contain between 1 and 50 characters, inclusive.
-
Each element of disk will be a single-space delimited list of integers, without extraneous leading zeros.
-
Each integer in disk will be between 0 and size-1, inclusive.
-
No integer will be appear more than once in disk.
Examples
0)

????
{"3 4 5 6 8 9 10","17 16 15"}
20
Returns: 5
We can defrag the first file by moving the contents of sector 8 to sector 7, then 9 to 8, and finally 10 to 9. The second file can be defragged in a number of ways by moving the contents of two sectors, for a total of 5.
1)

????
{"1 2 3 5 4 6 7 8"}
10
Returns: 2
Here we can take advantage of the fact that we have enough memory to hold two sectors worth of data. First, load the contents of sectors 4 and 5 into memory. Now, simply write the data back in the reverse order.
2)

????
{"1 3 5 7","0 2 4 8","6 9"}
100
Returns: 7

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

出处:蓝色理想
责任编辑:蓝色

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本文现有 4 条评论 暂时没有人参与评分


coverer Publish at 2005-10-14 10:45:32
一点都看不懂的啦!

www.album-frame.com

qiushuiwuhen Publish at 2005-10-11 15:21:56
要找到一个可以衡量的度,如所有不符合规律的细胞数和临界点的差值总合为0
先确定整数,当临界点的整数部分确定后,不符合规律的细胞也确定了,从而成为常数
((a-x)^2+(b-x)^2+...)/n(其中a,b,...,n都是常数,求导后,拐点就在差值总合为0)

++n;//记下所有不符合规律的细胞数
e+=(s[i]-th); //取得差值之和,有正有负,中和ing

th+=e/n; //把闸值修正,本来是整数
lovelement Publish at 2005-10-7 0:10:04
怎么第2个程序看不太懂 哪位大侠能帮解释一下啊
for(var th=nStart;nEnd;++th)
    {
        var e = 0,n=0;
        for(var i=0;i<s.length;++i)
        {
            if(c.charAt(i)!=(s[i]>=th?"C":"N"))
            {
                ++n;
                e+=(s[i]-th);
            }
        }
        if(0==n) return 0;
        if(e<0)
        {
            th+=e/n;
            break;
        }
    }
八神奄 Publish at 2005-8-23 16:56:24
标题有点学Csdn.... 其它的便不说了

一句中学生,再没有其它的描述吗? 哪个国家的,男的,女的? 简介;
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