Jack and Jill developed a special encryption method, so they can enjoy conversations without worrrying about eavesdroppers. Here is how: let L be the length of the original message, and M be the smallest square number greater than or equal to L. Add (M − L) asterisks to the message, giving a padded message with length M. Use the padded message to fill a table of size K × K, where K2= M. Fill the table in row-major order (top to bottom row, left to right column in each row). Rotate the table 90 degrees clockwise. The encrypted message comes from reading the message in row-major order from the rotated table, omitting any asterisks.

For example, given the
original message ‘iloveyouJack’, the message length is L = 12. Thus the padded
message is ‘iloveyouJack****’, with length M = 16. Below are the two tables
before and after rotation.

Then we read the secret message as
‘Jeiaylcookuv’.

The first line of input is the
number of original messages, 1 ≤ N ≤ 100. The following N lines each have a
message to encrypt. Each message contains only characters a–z (lower and upper
case), and has length 1 ≤ L ≤ 10 000.

For each original message,
output the secret message.

2

iloveyoutooJill

TheContestisOver

iteiloylloooJuv

OsoTvtnheiterseC

解题心得：
　　题意是输入字符串（长度L），然后将每个字符从上到下，从左到右挨着放到一个k*k的方格里，剩余的用*补上，然后将方格顺时针旋转一下（90度），然后在按照从上到下，从左到右的顺序输出（*省略掉）。K值：先把L开方，然后向上取整再加1，然后平方，即得到K的值。
　　我在做的时候忘记了把X清零，结果又是浪费了时间。。。

代码：

 #include <iostream>

 #include <cstring>

 #include <cstdio>

 #include <cmath>

 using namespace std;

 int main()

 {

     int n;

     int x=;

     int size=;

     double k=;

     int k1=;

     int s;

     char a[];

     char c[][];

     scanf("%d",&n);

     for(int i=;i<n;i++){

         cin>>a;

         size=strlen(a);

         k=sqrt(size);

         k1=ceil(k);

         s=k1*k1;

         if(size!=s){

             int add=s-size;

             for(int i1=;i1<add;i1++){

                 a[size+i1]='*';

             }

             a[size+add]='\n';

         }

          for(int j=;j<k1;j++){

                 for(int j1=;j1<k1;j1++){

                     c[j][j1]=a[x++];

                 }

             }

             x=; //做的时候被我落掉了，结果好久才找到错误！

             for(int r=;r<k1;r++){

                 for(int p=k1-;p>=;p--){

                     if(c[p][r]!='*'){

                         printf("%c",c[p][r]);

                     }

                 }

             }

             printf("\n");

     }

     return ;

 }