POJ2657Comfort（扩展欧几里得基础）

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

题意：

现在我们要找一个最小的a，使得男主从一号点出发，每次跳a格，中途不遇到障碍，最终走到Z点。

思路：

有方程 1+ax=ny+z。 最小的a需要保证x正整数解，且不存在s<x，使得1+as=ny+障碍的位置。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
int r[1010],n,z,m;
void Ex_gcd(int a,int b,int &d,int &x,int &y)
{
    if(b==0){x=1;y=0;d=a;return ;}
    Ex_gcd(b,a%b,d,y,x); y-=a/b*x;
}
bool check(int a)
{
    int d,x,y,X;
    Ex_gcd(a,n,d,X,y);
    if((z-1)%d!=0) return false;
    X=(((z-1)/d*X)%(n/d)+n/d)%(n/d);
    for(int i=1;i<=m;i++){
        Ex_gcd(a,n,d,x,y);
        if((r[i]-1)%d!=0) continue;
        x=(((r[i]-1)/d*x)%(n/d)+n/d)%(n/d);
        if(x<=X) return false;
    } return true;
}
int main()
{
    while(~scanf("%d%d%d",&n,&z,&m)){
        for(int i=1;i<=m;i++) scanf("%d",&r[i]);
        for(int i=1;i<n;i++) 
          if(check(i)){
             printf("%d\n",i); break;
          }
    } return 0;
}