SOJ 1007

1007. To and Fro

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

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 y h p k n n e l e a i r a h s g e c o n h s e m o t n 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

theresnoplacelikehomeonasnowynightx
thisistheeasyoneab
题目本身很简单，最大的难点应该是看懂英文题目，能看懂题的应该都会做……
// Problem#: 1007// Submission#: 4924383// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include<iostream>#include<string>using namespace std;int n;string A;int size;char **p;//用一个二维数组来保存字母 int pos;int flag=0;int main(){    while(cin>>n&&n)    {        pos=0;        flag=0;        cin>>A;        size=A.size();        p=new char*[size/n];        for(int i=0;i<size/n;++i)        p[i]=new char[n];        for(int i=0;i<size/n;++i)//由于生成密码时取字母是按s型取得，所以将字符串存回数组需要交替使用以下两个循环         {            if(!flag)//用flag来标记使用哪个循环，每次循环完成后记得改变flag的值以便下次进行另一个循环             for(int j=0;j<n;++j)            {                p[i][j]=A.at(pos);                pos++;                            }            else            for(int j=n-1;j>-1;--j)            {                p[i][j]=A.at(pos);                pos++;            }            if(flag)            flag=0;            else            flag=1;        }        for(int i=0;i<n;++i)        {            for(int j=0;j<size/n;++j)            cout<<p[j][i];        }        cout<<endl;        for(int i=0;i<size/n;++i)        delete []p[i];        delete []p;                    }    return 0; }