poj 2039 To and Fro

To and Fro

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 8460 Accepted: 5665

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

题意：给你个字符串和一个数字，把字符串写成如下规则：

对于There’s no place like home on a snowy night

有：

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

最后多余的用别的字母代替；

如何一列一列的输出即可；

#include <iostream>#include <string.h>using namespace std;int main(){int n;char s[500],map[500][500];while (cin>>n&&n){cin>>s;int len=strlen(s),k=0,flag=0;while (k<len){for (int i=0;i<n;i++)map[flag][i]=s[k++];flag++;if (k==len)break;for (int j=n-1;j>=0;j--)map[flag][j]=s[k++];flag++;}for (int i=0;i<n;i++){for (int j=0;j<flag;j++){cout<<map[j][i];}}cout<<endl;}return 0;}