# 杰拉斯的博客

## [ACM学习心得]关于sync_with_stdio(false);

杰拉斯 | 时间：2012-03-27, Tue | 23,840 views**编程算法**

在网上查看别人的ACM代码时，发现别人输入输出语句用的总是scanf与printf，有点不解，还以为他们用的都是C语言，而非C++，但今天做的一道题（Sort）：

发现与网上的其他高手使用完全相同的方法，使用scanf及printf的代码提交后Accepted，而使用cin及cout的却Time Limit Exceeded

## [整理]ACM详解（9）——其他

杰拉斯 | 时间：2012-02-17, Fri | 5,865 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。

**1、IP 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:

2

^{7}2^{6}2^{5}2^{4}2^{3}2^{2}2^{1}2^{0}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.

## [整理]ACM详解（8）——加密

杰拉斯 | 时间：2012-02-17, Fri | 6,686 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。

**1、Message 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.

## [整理]ACM详解（7）——压缩与编码

杰拉斯 | 时间：2012-02-17, Fri | 6,651 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。

**1、Parencodings**

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.

## [整理]ACM详解（6）——栈

杰拉斯 | 时间：2012-02-17, Fri | 6,858 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.