[暑假集训--数论]poj2657 Comfort

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

问在长度为n的环上走，每一次走k步，最后要走到z。有m个点是不可走的，问最小的k是多少

用exgcd可以解方程ax==b(mod c)，把这个式子写成ax-cy==b，exgcd解出ax+cy==gcd(a,c)，然后调一下系数，就能知道最小的x。

如果0到z-1的步数大于了0到某一个障碍位置的步数，说明先到障碍位置，就不行

 #include<cstdio>
 #include<iostream>
 #define LL long long
 using namespace std;
 inline LL read()
 {
     LL x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 int n,z,m;
 int b[];
 inline int exgcd(int a,int b,int &x,int &y)
 {
     if (!b){x=;y=;return a;}
     int gcd=exgcd(b,a%b,x,y);
     int t=x;x=y;y=t-a/b*y;
     return gcd;
 }
 inline int calc(int a,int b,int c)//a*x==b(mod c)
 {
     int x=,y=;
     int tt=exgcd(a,c,x,y);
     if (b%tt!=)return -;x=(x*b/tt)%c;
     int ss=c/tt;
     x=(x%ss+ss)%ss;
     return x;
 }
 int main()
 {
     while (~scanf("%d%d%d",&n,&z,&m))
     {
         z--;
         for(int i=;i<=m;i++)
             b[i]=read()-;
         for (int i=;i<=z;i++)
         {
             bool ok=;
             int step=calc(i,z,n);
             if (step==-)continue;
             for (int j=;j<=m;j++)
             {
                 int now=calc(i,b[j],n);
                 if (now==-||now>step)continue;
                 ok=;break;
             }
             if (ok){printf("%d\n",i);break;}
         }
     }
 }

poj 2657