归档:2012年2月月

[整理]ACM详解(9)——其他

蓝飞 蓝飞 | 时间:2012-02-17, Fri | 4,055 views
编程算法 
有时候会考一些锻炼基本能力的题目,下面使用几个例子进行简单分析。
1IP Address
Description
Suppose you are reading byte streams from any device, representing IP addresses. Your task is to convert a 32 characters long sequence of '1s' and '0s' (bits) to a dotted decimal format. A dotted decimal format for an IP address is form by grouping 8 bits at a time and converting the binary representation to decimal representation. Any 8 bits is a valid part of an IP address. To convert binary numbers to decimal numbers remember that both are positional numerical systems, where the first 8 positions of the binary systems are:
27 26 25 24 23 22 21 20
128  64 32 16 8   4   2   1
Input
The input will have a number N (1<=N<=9) in its first line representing the number of streams to convert. N lines will follow.
Output
The output must have N lines with a doted decimal IP address. A dotted decimal IP address is formed by grouping 8 bit at the time and converting the binary representation to decimal representation.

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[整理]ACM详解(8)——加密

蓝飞 蓝飞 | 时间:2012-02-17, Fri | 4,254 views
编程算法 
比赛的时候告诉你简单的加密算法,让你完成加密或者解密操作,下面通过几个简单例子介绍。
1Message Decowding
Description
The cows are thrilled because they've just learned about encrypting messages. They think they will be able to use secret messages to plot meetings with cows on other farms.
Cows are not known for their intelligence. Their encryption method is nothing like DES or BlowFish or any of those really good secret coding methods. No, they are using a simple substitution cipher.
The cows have a decryption key and a secret message. Help them decode it. The key looks like this:
yrwhsoujgcxqbativndfezmlpk
Which means that an 'a' in the secret message really means 'y'; a 'b' in the secret message really means 'r'; a 'c' decrypts to 'w'; and so on. Blanks are not encrypted; they are simply kept in place.
Input text is in upper or lower case, both decrypt using the same decryption key, keeping the appropriate case, of course.
Input
* Line 1: 26 lower case characters representing the decryption key
* Line 2: As many as 80 characters that are the message to be decoded
Output
* Line 1: A single line that is the decoded message. It should have the same length as the second line of input.

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[整理]ACM详解(7)——压缩与编码

蓝飞 蓝飞 | 时间:2012-02-17, Fri | 4,824 views
编程算法 

有些题目会给出一些简单的压缩方法或者编码方法,让你实现具体的算法。下面通过题目分析。

1Parencodings
Description
Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways:
By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).
By an integer sequence W = w1 w2...wn where for each right parenthesis, say a in S, we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a. (W-sequence).
Following is an example of the above encodings:
S                 (((()()())))
P-sequence          4 5 6666
W-sequence         1 1 1456
Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
Input
The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case is an integer n (1 <= n <= 20), and the second line is the P-sequence of a well-formed string. It contains n positive integers, separated with blanks, representing the P-sequence.
Output
The output file consists of exactly t lines corresponding to test cases. For each test case, the output line should contain n integers describing the W-sequence of the string corresponding to its given P-sequence.

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[整理]ACM详解(6)——栈

蓝飞 蓝飞 | 时间:2012-02-17, Fri | 5,038 views
编程算法 
堆栈是一种特殊的线性结构,后进先出,只能对栈顶元素操作,典型的操作入栈和出站。下面通过例子介绍基本用法。
题目:

Train Problem

Problem Description
As the new term comes, the Ignatius Train Station is very busy nowadays. A lot of student want to get back to school by train(because the trains in the Ignatius Train Station is the fastest all over the world ^v^). But here comes a problem, there is only one railway where all the trains stop. So all the trains come in from one side and get out from the other side. For this problem, if train A gets into the railway first, and then train B gets into the railway before train A leaves, train A can't leave until train B leaves. The pictures below figure out the problem. Now the problem for you is, there are at most 9 trains in the station, all the trains has an ID(numbered from 1 to n), the trains get into the railway in an order O1, your task is to determine whether the trains can get out in an order O2.
Input
The input contains several test cases. Each test case consists of an integer, the number of trains, and two strings, the order of the trains come in:O1, and the order of the trains leave:O2. The input is terminated by the end of file. More details in the Sample Input.
Output
The output contains a string "No." if you can't exchange O2 to O1, or you should output a line contains "Yes.", and then output your way in exchanging the order(you should output "in" for a train getting into the railway, and "out" for a train getting out of the railway). Print a line contains "FINISH" after each test case. More details in the Sample Output.

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[整理]ACM详解(5)——排序

蓝飞 蓝飞 | 时间:2012-02-17, Fri | 3,474 views
编程算法 
有些ACM题需要使用一些基本的数据结构,下面首先介绍与排序相关的内容。
1、基本排序
Problem Description
These days, I am thinking about a question, how can I get a problem as easy as A+B? It is fairly difficulty to do such a thing. Of course, I got it after many waking nights.
Give you some integers, your task is to sort these number ascending (升序).
You should know how easy the problem is now!
Good luck!
Input
Input contains multiple test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. Each test case contains an integer N (1<=N<=1000 the number of integers to be sorted) and then N integers follow in the same line.
It is guarantied that all integers are in the range of 32-int.
Output
For each case, print the sorting result, and one line one case.

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