Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 9285		Accepted: 6261

Description

Mo and Larry have devised a way of encrypting messages. They first decide secretly on the number of columns and write the message (letters only) down the columns, padding with extra random letters so as to make a rectangular array of letters. For example, if the message is "There’s no place like home on a snowy night" and there are five columns, Mo would write down  

t o i o yh p k n ne l e a ir a h s ge c o n hs e m o tn l e w x

Note that Mo includes only letters and writes them all in lower case. In this example, Mo used the character "x" to pad the message out to make a rectangle, although he could have used any letter.  

Mo then sends the message to Larry by writing the letters in each row, alternating left-to-right and right-to-left. So, the above would be encrypted as  

toioynnkpheleaigshareconhtomesnlewx  

Your job is to recover for Larry the original message (along with any extra padding letters) from the encrypted one.  

Input

There will be multiple input sets. Input for each set will consist of two lines. The first line will contain an integer in the range 2. . . 20 indicating the number of columns used. The next line is a string of up to 200 lower case letters. The last input set is followed by a line containing a single 0, indicating end of input.

Output

Each input set should generate one line of output, giving the original plaintext message, with no spaces.

Sample Input

5toioynnkpheleaigshareconhtomesnlewx3ttyohhieneesiaabss0

Sample Output

theresnoplacelikehomeonasnowynightxthisistheeasyoneab

Source

Regionals 2004 >> North America - East Central NA

问题链接：。

问题简述：（略）

问题分析：

虽然字符串是放在一维数组里，也可以按二维数组的形式对其进行访问，是一个下标映射的问题。

此外，还需要考虑行的奇偶和行中的顺序访问（奇数行顺序访问）与逆序访问（偶数行逆序访问）。

程序说明：（略）

AC的C语言程序（代码中有C++的注释，需要用C++编译）如下：

/* HDU1200 POJ2039 ZOJ2208 UVALive3084 To and Fro */#include 
    
     #include 
     
      int main(void){    int c, r, oddeven, len, i, j;    char s[200+1];    while(scanf("%d", &c) != EOF) {        if(c == 0)            break;        scanf("%s", s);        len = strlen(s);        r = len / c;        for(j=0; j