POJ2013 ZOJ2172 UVALive3055 Symmetric Order【递归+堆栈】

xiaoxiao2021-02-28  34

Symmetric Order Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 14431 Accepted: 8559

Description

In your job at Albatross Circus Management (yes, it's run by a bunch of clowns), you have just finished writing a program whose output is a list of names in nondescending order by length (so that each name is at least as long as the one preceding it). However, your boss does not like the way the output looks, and instead wants the output to appear more symmetric, with the shorter strings at the top and bottom and the longer strings in the middle. His rule is that each pair of names belongs on opposite ends of the list, and the first name in the pair is always in the top part of the list. In the first example set below, Bo and Pat are the first pair, Jean and Kevin the second pair, etc.  

Input

The input consists of one or more sets of strings, followed by a final line containing only the value 0. Each set starts with a line containing an integer, n, which is the number of strings in the set, followed by n strings, one per line, sorted in nondescending order by length. None of the strings contain spaces. There is at least one and no more than 15 strings per set. Each string is at most 25 characters long.

Output

For each input set print "SET n" on a line, where n starts at 1, followed by the output set as shown in the sample output.

Sample Input

7 Bo Pat Jean Kevin Claude William Marybeth 6 Jim Ben Zoe Joey Frederick Annabelle 5 John Bill Fran Stan Cece 0

Sample Output

SET 1 Bo Jean Claude Marybeth William Kevin Pat SET 2 Jim Zoe Frederick Annabelle Joey Ben SET 3 John Fran Cece Stan Bill

Source

Mid-Central USA 2004

Regionals 2004 >> North America - Mid-Central USA

问题链接:POJ2013 ZOJ2172 UVALive3055 Symmetric Order

问题简述:(略)

问题分析:

  这个问题是要把输入的字符串改变一个顺序输出,使用递归来实现则比较方便。递归函数把先输入后输出的的字符串暂时保存在变量中,等处理完其他的字符串,再输出先输入的字符串。

  可以用递归实现的功能,往往也可以用堆栈来实现。

程序说明:(略)

  给出两个程序,一个是用堆栈实现,另外一个是用递归实现。

题记:(略)

参考链接:(略)

AC的C++语言程序如下(堆栈):

/* POJ2013 ZOJ2172 UVALive3055 Symmetric Order */ #include <iostream> #include <string> #include <stack> using namespace std; int main() { int n, caseno = 0; string s; while(cin >> n && n) { stack<string> stk; cout << "SET " << ++caseno << endl; while(n) { cin >> s; cout << s << endl; if(--n) { cin >> s; stk.push(s); n--; } } while(!stk.empty()) { cout << stk.top() << endl; stk.pop(); } } return 0; }

AC的C++语言程序如下(递归):

/* POJ2013 ZOJ2172 UVALive3055 Symmetric Order */ #include <iostream> using namespace std; void print(int n) { string s; cin >> s; cout << s << endl; if(--n) { cin >> s; if(--n) print(n); cout << s << endl; } } int main() { int n, caseno = 0; while(cin >> n && n) { cout << "SET " << ++caseno << endl; print(n); } return 0; }

转载请注明原文地址: https://www.6miu.com/read-1650150.html

最新回复(0)