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 ��wrap around�� 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 illustrated 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 should not be removed, they are considered part of the number and count in determining n. Thus, ��01�� is a two-digit number, distinct from ��1�� which 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 

/*

雖說是水題，但是考察的知識點也是很坑爹的，這個數如果是迴圈數則這個數乘以這個數的長度+1 是一個各位全是9的數

*/

#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;}