Runaround numbers are integers with unique digits, none of which is zero (e.g., 81362) that also have an interesting property, exemplified by this demonstration:

• If you start at the left digit (8 in our number) and count that number of digits to the right (wrapping back to the first digit when no digits on the right are available), you'll end up at a new digit (a number which does not end up at a new digit is not a Runaround Number). Consider: 8 1 3 6 2 which cycles through eight digits: 1 3 6 2 8 1 3 6 so the next digit is 6.
• Repeat this cycle (this time for the six counts designed by the `6') and you should end on a new digit: 2 8 1 3 6 2, namely 2.
• Repeat again (two digits this time): 8 1
• Continue again (one digit this time): 3
• One more time: 6 2 8 and you have ended up back where you started, after touching each digit once. If you don't end up back where you started after touching each digit once, your number is not a Runaround number.

Given a number M (that has anywhere from 1 through 9 digits), find and print the next runaround number higher than M, which will always fit into an unsigned long integer for the given test data.

PROGRAM NAME: runround

INPUT FORMAT

SAMPLE INPUT (file runround.in)

81361

OUTPUT FORMAT

SAMPLE OUTPUT (file runround.out)

81362

題意：

例如對81362,默認取第一個數‘8',向后數8位，如果位數不夠8個，則繼續從頭開始數起，直到‘6’，接著是‘2’,再接著
是‘1’，再到‘3’,這樣所有數都visit過了。接著再到‘8’,能回到起點（必須要回到起點）。在滿足以上條件下，要求
每位上的數是唯一的且不含0，這樣的一個數就是runround number。

對給出的n，求出n后面得第一個runround number。

代碼如下：

/*
LANG: C
TASK: runround
*/
#include<stdio.h>
int s[20];
int T[10];
int turn(int n)
{
    int i = 0;
    while (n)
    {
        s[i] = n % 10;
        T[s[i]] = 0;//初始化標記數組T 
        i++;
        n /= 10;
    }
    return i;
}
int Ok(int len)
{
    int i, pre, r, local;
    if (len == 0)
    {
        return 0;
    }
    for (i = len - 1; i >= 0; i--)
    {
        if (s[i] == 0 || T[s[i]] == 1)//判0和判是否存在數重復 
        {
            return 0;
        }
        T[s[i]] = 1;//對掃過的數標記 
    }
    for (i = len - 1; i >= 0; i--)
    {
        T[s[i]] =  0;
    }
    pre = s[len-1];
    T[pre] = 1;
    r = 1;
    local = len - 1;
    while (1)
    {
        pre %= len;
        //尋找下一個數的下標 
        if (local - pre >= 0)//如果下一個數在pre后面，及下標小于pre的下標 
        {
            local =  local - pre;
        }
        else
        {
            local = len + (local - pre);//如果下一個數在pre前面，即下標大于pre的下標 
        }
        pre = s[local];
        if (r == len)//全部掃描過 
        {
            if (pre == s[len-1])//如果能回到起點 
            {
                return 1;
            }
            else
            {
                return 0;
            }
        }
        else if (r != len && T[pre])//在沒掃描所有數就返回到被標記過的數，則一定不能返回起點 
        {
            return 0;
        }        
        T[pre] = 1;//標記掃描過的數 
        r++;//掃描個數加1    
    }                 
}
int main()
{
    freopen("runround.in", "r", stdin);
    freopen("runround.out", "w", stdout);
    int n, len, t, i;
    scanf("%d", &n);
    for (i = n + 1; ; i++)
    {
        len = turn(i);
        t = Ok(len);
        if (t)
        {
            break;
        }
    }
    printf("%d\n", i);
    fclose(stdin);
    fclose(stdout);
    //system("pause");
    return 0;    
}