CIPHERJ-Cipher

Cipher

below given code is for cipherj spoj or cipher spoj.

In this problem you have to find cycle of repetition .

lets take given example.

10
4 5 3 7 2 8 1 6 10 9
1 Hello Bob
1995 CERC
0
0

for given array of ten length repetition are as follows.

___|0__1__2__3__4__5__6__7__8__9__10_

   |

 0 |0  1  2  3  4  5  6  7  8  9  10

 1 |0  4  5     7  2  8  1  6  10  9

 3 |0  7        1        4          

word in first position will repeat on 1st 4th and 7th position so if given k=1995

then word at first position will go-to pos=1995%3 which is 0 so number at index [0,1]=1 character at position 1 is C;

and so on for all number;

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <cstdlib>
using namespace std;
int main()
{
 int n;
 while(1)
 {
  scanf("%d",&n);
  if(n==0)
   break;
  int a[n+5];
  int i,j,k,count=1;
  int res[210][210];
  for(i=1;i<=n;i++)
  {
   scanf("%d",&a[i]);
   res[0][i]=i;
  }
  while(1)
  {
   scanf("%d",&k);
   if(k==0)
    break;
   char *word;
   word=(char *)calloc(210,sizeof(char));
   char *final=(char *)calloc(210,sizeof(char));
   fflush(stdin);
   gets(word);
   int count,pos;
   for(i=1;i<=n;i++)
   {
    count=1;
    j=i;
    while(a[j]!=i)
    {
     res[count++][i]=a[j];
     j=a[j];
    }
    pos=res[k%count][i];
    final[pos-1]=word[i]==0?32:word[i];
   }
   for(j=0;j<n;j++)
    printf("%c",final[j]);
    printf("\n");
  }
 }
 return 0;
}