Round and Round We Go

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10514		Accepted: 4812

Description

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: 142857 *1 = 142857 142857 *2 = 285714 142857 *3 = 428571 142857 *4 = 571428

142857 *5 = 714285 142857 *6 = 857142

Input

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.

Sample Input

142857142856142858010588235294117647

Sample Output

142857 is cyclic142856 is not cyclic142858 is not cyclic01 is not cyclic0588235294117647 is cyclic

#include 
     
      #include 
      
       #include 
       
        using namespace std;const int N = 65;char str[N];int num[N],ans[N],len;int cmp(const void *a,const void *b){    return *(int *)a-*(int *)b;}bool is_match(){    qsort(num,len,sizeof(int),cmp);    qsort(ans,len,sizeof(int),cmp);    int i;    for(i=0;i
        
         =0;i--)             num[j++]=str[i]-'0';        for(i=2;i<=len;i++)        {            memset(ans,0,sizeof(ans));            for(j=0;j
         
          9)                {                    ans[k+1] += ans[k]/10;//两条语句不可交换，搞了好久才发现                     ans[k] %= 10;                }            }            bool flag = is_match();            if(flag)            {                if(i==len)                {                    for(j=0;j
          
           =0;j--) num[k++]=str[j]-'0'; } else { for(j=0;j
          
         
        
       
      
     

 

对于数N，若N为循环的则有N*(length(N)+1)=99....99， (length(N)个9)，length(N)为N的位数，含前导0。应该还是简单的...   

//此为转载

#include<iostream>

#include<string>

using

namespace

std;

bool

fun(string str)

{

   

int

n=str.length()+1;

   

int

i,up=0,temp=0;

   

for

(i=n-2;i>=0;i--)

   

{

       

temp=(

int

)(str[i]-

'0'

);

       

if

((temp*n+up)%10!=9)

return

false

;

       

up=(temp*n+up)/10;

   

}

   

return

true

;

}

int

main()

{

   

string str1;

   

while

(cin>>str1)

   

{

       

if

(fun(str1))

       

{

           

cout<<str1<<

" is cyclic"

<<endl;

       

}

       

else

       

{

           

cout<<str1<<

" is not cyclic"

<<endl;

       

}

   

}

   

return

0;

}