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[65];
int num[65];
int temp[65];
int n;

bool cacl(int m)
{
    int t = 0;
    for (int i = 0; i < n; i++)
    {
        temp[i] = (num[i] * m + t) % 10;
        t = (num[i] * m + t) / 10;
    }
    if (t>0)  return false;
    return true;
}

bool judge()
{
    for (int i = 0; i<n; ++i)
    {
        int k = 0;
        if (temp[i] == num[0])
        {
             while (k<n &&num[++k] == temp[(i+k) % n]);
             if (k == n)   return 1;   
        }
    }
    return 0;
}

int main()
{
    //freopen("D:\txt.txt", "r", stdin);
    while (gets(s))
    {
        memset(num, 0, sizeof(num));
        n = strlen(s);
        for (int i = n - 1; i >= 0; i--)
        {
            num[n - 1 - i] = s[i] - '0';
        }
        bool flag = false;
        for (int i = 2; i <= n; i++)
        {
            if (cacl(i))
            {
                if (!judge())
                {
                    printf("%s is not cyclic
", s);
                    flag = true;
                    break;
                }
            }
            else
            {
                printf("%s is not cyclic
", s);
                flag = true;
                break;
            }
        }
        if (!flag)
            printf("%s is cyclic
", s);
    }
    return 0;
}