POJ-1047 Round and Round We Go

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Round and Round We Go

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10483		Accepted: 4802

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

142857
142856
142858
01
0588235294117647

Sample Output

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

 1 /* 
 2    功能Function Description:     POJ-1047 大数乘法 + 匹配 
 3    开发环境Environment:          DEV C++ 4.9.9.1
 4    技术特点Technique:
 5    版本Version:
 6    作者Author:                   可笑痴狂
 7    日期Date:                      20120803
 8    备注Notes:
 9 */
10 #include<stdio.h>
11 #include<string.h>
12 int len;
13 
14 int mult(int *num,int *temp,int n) //num中的数与n相乘存到temp中，如果结果位数大于num则返回0，否则返回1
15 {
16     int i,t=0;
17     for(i=0;i<len;++i)
18     {
19         t=num[i]*n+t;
20         temp[i]=t%10;
21         t/=10;
22     }
23     if(t)
24         return 0;
25     return 1;
26 }
27 
28 int Judge(int *num,int *temp)    //判断是否匹配
29 {
30     int i,j,k;
31     for(i=0;i<len;++i)
32     {
33         k=0;
34         if(temp[i]==num[0])
35         {
36             j=i;
37             while(k<len&&num[++k]==temp[(++j)%len]);
38             if(k==len)
39                 return 1;   //说明可以匹配
40         }
41     }
42     return 0;
43 }
44 
45 int main()
46 {
47     char s[65];
48     int num[65];
49     int temp[65];
50     int i,j,flag;
51     while(gets(s))
52     {
53         flag=1;
54         len=strlen(s);
55         memset(num,0,sizeof(num));
56         for(i=len-1,j=0;i>=0;--i)
57             num[j++]=s[i]-'0';
58         for(i=2;i<=len;++i)
59         {
60             memset(temp,0,sizeof(temp));
61             if(mult(num,temp,i))    
62             {
63                 if(Judge(num,temp)==0)
64                 {
65                     printf("%s is not cyclic\n",s);
66                     flag=0;
67                     break;
68                 }
69             }
70             else //相乘的结果位数增多肯定不满足，直接跳出
71             {
72                 printf("%s is not cyclic\n",s);
73                 flag=0;
74                 break;
75             }
76 
77         }
78         if(flag)
79             printf("%s is cyclic\n",s);
80     }
81     return 0;
82 }