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.
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.
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
1 #include<cstdio> 2 #include<iostream> 3 #define LL long long 4 using namespace std; 5 inline LL read() 6 { 7 LL x=0,f=1;char ch=getchar(); 8 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 9 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 10 return x*f; 11 } 12 int n,z,m; 13 int b[100010]; 14 inline int exgcd(int a,int b,int &x,int &y) 15 { 16 if (!b){x=1;y=0;return a;} 17 int gcd=exgcd(b,a%b,x,y); 18 int t=x;x=y;y=t-a/b*y; 19 return gcd; 20 } 21 inline int calc(int a,int b,int c)//a*x==b(mod c) 22 { 23 int x=0,y=0; 24 int tt=exgcd(a,c,x,y); 25 if (b%tt!=0)return -1;x=(x*b/tt)%c; 26 int ss=c/tt; 27 x=(x%ss+ss)%ss; 28 return x; 29 } 30 int main() 31 { 32 while (~scanf("%d%d%d",&n,&z,&m)) 33 { 34 z--; 35 for(int i=1;i<=m;i++) 36 b[i]=read()-1; 37 for (int i=1;i<=z;i++) 38 { 39 bool ok=1; 40 int step=calc(i,z,n); 41 if (step==-1)continue; 42 for (int j=1;j<=m;j++) 43 { 44 int now=calc(i,b[j],n); 45 if (now==-1||now>step)continue; 46 ok=0;break; 47 } 48 if (ok){printf("%d ",i);break;} 49 } 50 } 51 }