USACO 2.2.3 Runaround Numbers

题意：

求比给出的M大的循环数， 循环数，从位置0开始数arr[0]个数，然后再从位置i数arr[i]个数，最后回到起点

解法：

简单模拟

/*
ID: lsswxr1
PROG: runround
LANG: C++
*/
#include <iostream>
#include <vector>
#include <map>
#include <list>
#include <set>
#include <deque>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <cstdio>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <fstream>
using namespace std;

#define USACO
#ifdef USACO
#define cin fin
#define cout fout
#endif
//////////////////////////////////////////////////////////////////////////
///宏定义
const int  INF = 1000000000;
const int MAXN = 40001;
const int maxn = MAXN;
///全局变量 和 函数
long long N;
int nums[30];
bool vis[30];
bool hasNum[10];
bool isok(int cur)
{
    memset(hasNum, false, sizeof(hasNum));
    int cnt = 0;
    int temp[30];
    int divNum = 10;
    int Number = cur;
    while (Number != 0)
    {
        int saveNum = Number % 10;
        if (saveNum == 0 || hasNum[saveNum] == true)
            return false;
        hasNum[saveNum] = true;
        temp[cnt++] = saveNum;
//        divNum *= 10;
        Number /= divNum;
    }
    int j = 0;
    for (int i = cnt - 1; i >= 0; i--)
    {
        nums[j++] = temp[i];
    }

    memset(vis, false, sizeof(vis));
    int counter = 0;
    int startNum = nums[0];
    int startPos = 0;
    vis[0] = true;
    counter++;
    while (counter < cnt)
    {
        startPos = (startPos + startNum) % cnt;
        if (vis[startPos])
        {
            return false;
        }
        else
        {
            vis[startPos] = true;
            startNum = nums[startPos];
            counter++;
        }
    }
    int results = (startPos + startNum) % cnt;
    if (results == 0)
        return true;
    return false;
}


int main()
{


#ifdef USACO    
    ofstream fout ("runround.out");
    ifstream fin ("runround.in");
#endif   
     while (cin >> N)
     {
         int M = N + 1;
         while (1)
         {
             if (isok(M))
             {
                 cout << M << endl;
                 break;
             }
             M++;
         }
     }
    
    
    ///结束
    return 0;
}