POJ 1047 Round and Round We Go

https://vjudge.net/problem/POJ-1047

题意：

给一个整数，它的长度为n，从1开始一直到n和该整数相乘，判断每次结果是否和原来的整数是循环的。

思路：

大整数的乘法。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include<vector>
 7 using namespace std;
 8
 9 char s[65];
10 int num[65];
11 int temp[65];
12 int n;
13
14 bool cacl(int m)
15 {
16     int t = 0;
17     for (int i = 0; i < n; i++)
18     {
19         temp[i] = (num[i] * m + t) % 10;
20         t = (num[i] * m + t) / 10;
21     }
22     if (t>0)  return false;
23     return true;
24 }
25
26 bool judge()
27 {
28     for (int i = 0; i<n; ++i)
29     {
30         int k = 0;
31         if (temp[i] == num[0])
32         {
33              while (k<n &&num[++k] == temp[(i+k) % n]);
34              if (k == n)   return 1;
35         }
36     }
37     return 0;
38 }
39
40 int main()
41 {
42     //freopen("D:\\txt.txt", "r", stdin);
43     while (gets(s))
44     {
45         memset(num, 0, sizeof(num));
46         n = strlen(s);
47         for (int i = n - 1; i >= 0; i--)
48         {
49             num[n - 1 - i] = s[i] - ‘0‘;
50         }
51         bool flag = false;
52         for (int i = 2; i <= n; i++)
53         {
54             if (cacl(i))
55             {
56                 if (!judge())
57                 {
58                     printf("%s is not cyclic\n", s);
59                     flag = true;
60                     break;
61                 }
62             }
63             else
64             {
65                 printf("%s is not cyclic\n", s);
66                 flag = true;
67                 break;
68             }
69         }
70         if (!flag)
71             printf("%s is cyclic\n", s);
72     }
73     return 0;
74 }