[模拟]Comfort

Comfort

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2476		Accepted: 847

Description

A game-board consists of N fields placed around a circle. Fields are successively numbered from1 to N clockwise. In some fields there may be obstacles. 

Player starts on a field marked with number 1. His goal is to reach a given field marked with number Z. The only way of moving is a clockwise jump of length K. The only restriction is that the fields the player may jump to should not contain any obstacle. 

For example, if N = 13, K = 3 and Z = 9, the player can jump across the fields 1, 4, 7, 10, 13, 3, 6 and 9, reaching his goal under condition that none of these fields is occupied by an obstacle. 

Your task is to write a program that finds the smallest possible number K. 

Input

First line of the input consists of integers N, Z and M, 2 <= N <= 1000, 2 <= Z <= N, 0 <= M <= N - 2. N represents number of fields on the game-board and Z is a given goal-field. 

Next line consists of M different integers that represent marks of fields having an obstacle. It is confirmed that fields marked 1 and Z do not contain an obstacle.

Output

Output a line contains the requested number K described above.

Sample Input

9 7 2
2 3

Sample Output

3

Source

Croatia OI 2002 National – Juniors

注意如果直到出现循环仍然没有走到Z，则说明该K值不合法。

#include <cstdio>
#include <cstring>

bool stepped[1010];
bool obs[1010];

int main()
{
    freopen("2657.in","r",stdin);
    freopen("2657.out","w",stdout);

    int n,z,m;
    while (scanf("%d%d%d",&n,&z,&m) == 3)
    {
        memset(obs,0,sizeof obs);

        z --;
        for (int i=1;i<=m;i++)
        {
            int tmp;
            scanf("%d",&tmp);
            tmp --;
            obs[tmp] = true;
        }

        bool ok = false;
        for (int t=1;t<=z&&!ok;t++)
        {
            for (int i=0;i<n;i++)
                stepped[i] = false;
            for (int ii=t;!stepped[ii%n]&&!obs[ii%n];ii+=t)
            {
                stepped[ii%n] = true;
                if (ii%n == z)
                {
                    ok = true;
                    printf("%d\n",t);
                    break;
                }
            }
        }
    }
    return 0;
}