Round and Round We Go

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 658    Accepted Submission(s): 365

Problem Description

A cyclic number is an integer n digits in length which, when multiplied by any integer from 1 to n, yields a ~{!0~}cycle~{!1~} of the digits of the original number. That is, if you consider the number after the last digit to ~{!0~}wrap around~{!1~} 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

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, ~{!0~}01~{!1~} is a two-digit number, distinct from ~{!0~}1~{!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

题意：

给出一个数（长度<=60），然后这个数依次乘1--n（n为该数的长度），若每次乘后所得的数首尾相连构成的环和给的那个数构成的环相等，则该数是环。

思路：

高精度模拟一下就行了。

代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <vector>
 5 #include <algorithm>
 6 using namespace std;
 7
 8 char s[100];
 9 char str[200];
10
11 int solve(){
12     int a[100];
13     char s1[100];
14     int i, j, n=strlen(s);
15     for(i=2;i<=n;i++){
16         memset(s1,‘\0‘,sizeof(s1));
17         memset(a,0,sizeof(a));
18         for(j=n-1;j>=0;j--) a[j]=s[n-j-1]-‘0‘;
19         for(j=0;j<n;j++) a[j]*=i;
20         for(i=0;i<=n+2;i++){
21             a[i+1]=a[i+1]+a[i]/10;
22             a[i]%=10;
23         }
24         int k=n+2;
25         while(!a[k]) k--;
26         for(j=n-1;j>=0;j--) s1[j]=a[n-j-1]+‘0‘;
27         if(!strstr(str,s1)) return 0;
28     }
29     return 1;
30 }
31
32 main()
33 {
34     while(scanf("%s",s)!=EOF){
35         strcpy(str,s);
36         strcat(str,s);
37         int ans=solve();
38         if(ans) printf("%s is cyclic\n",s);
39         else printf("%s is not cyclic\n",s);
40     }
41 }

HDU 1313 高精度*单精度,布布扣,bubuko.com