Time Limit:1000MS	Memory Limit:30000K
Total 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

問題簡述：（略）

問題分析：

　　這個問題是要把輸入的字串改變一個順序輸出，使用遞迴來實現則比較方便。遞迴函式把先輸入後輸出的的字串暫時儲存在變數中，等處理完其他的字串，再輸出先輸入的字串。

　　可以用遞迴實現的功能，往往也可以用堆疊來實現。

程式說明：（略）

　　給出兩個程式，一個是用堆疊實現，另外一個是用遞迴實現。

題記：（略）

參考連結：（略）

AC的C++語言程式如下（堆疊）：

/* 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++語言程式如下（遞迴）：

/* 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;
}