Zoj problem set-1073 Round and round we go ()

Problem

A cyclic Number is an integer n digits in length which, when multiplied by any integer from 1 to n, yields a �� cycle �� of the digits of the original number. that is, if you consider the number after the last digit to Fig back to the first digit, the sequence of digits in both numbers will be the same, though they may start at different positions.

For example, the number 142857 is cyclic, as partitioned strated by the following table:

Write a program which will determine whether or not numbers are cyclic. the input file is a list of integers from 2 to 60 digits in length. (Note that preceding zeros shocould not be removed, they are considered part of the number and count in determining n. thus, fig is a two-digit number, distinct from Fig is a one-digit number .)

Output

For each input integer, write a line in the output indicating whether or not it is cyclic.

 

Example

Input

142857
142856
142858
01
0588235294117647

Output

142857 is cyclic
142856 is not cyclic
142858 is not cyclic
01 is not cyclic
0588235294117647 is cyclic

/*

Although it is a question of water, the knowledge points of investigation are also very poor. If this number is a loop number, then multiply this number by the length of this number + 1 is a number all 9

*/

#include <iostream>#include <stdio.h>#include <memory>#include <string.h>using namespace std; const int maxn=70;int std_str[maxn];bool search(){}int main () { char str[maxn]; int ans[maxn],res[maxn]; memset(std_str,9,sizeof(std_str)); while ( memset(str,'/0',70),cin>>str) {  int len=strlen(str),i,j;  for ( i=0,j=len-1;i<len;j--,i++)   ans[i]=str[j]-48;  memset(res,0,sizeof(res)) ;  for ( i=0;i<len;i++)   ans[i]*=len+1;  for ( i=0;i<len;i++)  {   if( ans[i]>9)   {    ans[i+1]+=ans[i]/10;    ans[i]%=10;   }  }   int t=1;  for(i=0;i<len;i++)   if(ans[i]!=9){t=0;break;}  if(t)  cout<<str<<" is cyclic/n";  else cout <<str<<" is not cyclic/n" ; } return 0;}