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。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include<vector>
 using namespace std;

 char s[];
 int num[];
 int temp[];
 int n;

 bool cacl(int m)
 {
     int t = ;
     for (int i = ; i < n; i++)
     {
         temp[i] = (num[i] * m + t) % ;
         t = (num[i] * m + t) / ;
     }
     if (t>)  return false;
     return true;
 }

 bool judge()
 {
     for (int i = ; i<n; ++i)
     {
         int k = ;
         if (temp[i] == num[])
         {
              while (k<n &&num[++k] == temp[(i+k) % n]);
              if (k == n)   return ;
         }
     }
     return ;
 }

 int main()
 {
     //freopen("D:\\txt.txt", "r", stdin);
     while (gets(s))
     {
         memset(num, , sizeof(num));
         n = strlen(s);
         for (int i = n - ; i >= ; i--)
         {
             num[n -  - i] = s[i] - '';
         }
         bool flag = false;
         for (int i = ; i <= n; i++)
         {
             if (cacl(i))
             {
                 if (!judge())
                 {
                     printf("%s is not cyclic\n", s);
                     flag = true;
                     break;
                 }
             }
             else
             {
                 printf("%s is not cyclic\n", s);
                 flag = true;
                 break;
             }
         }
         if (!flag)
             printf("%s is cyclic\n", s);
     }
     return ;
 }