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.
i l o v * j e i
e y o y * a y l
j a c k * c o o
* * * * * k y v
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
#include<stdio.h>
#include"iostream"
#include<string.h>
#include"math.h"
using namespace std;
#define N 10005
int main()
{
char str2[N];
int i,j,k,t,f=0,s,l,m;
int n;
scanf("%d",&n);
getchar();
for(s=0;s<n;s++)
{
memset(str2,'*',10005*sizeof(char));
gets(str2);
l=strlen(str2);
m=ceil(sqrt(l));
k=m*m;
str2[l]='*';
str2[k]=' ';
for(j=k-m;j<k;j++)
{
f=j;
while(f>=0){
if(str2[f]!='*')
printf("%c",str2[f]);
f=f-m;
}
}
printf("
");
}
return 0;
}