A strange lift

题目链接：

题目：

There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button "UP" , you will go up Ki floor,i.e,you will go to the i+Ki th floor,as the same, if you press the button "DOWN" , you will go down Ki floor,i.e,you will go to the i-Ki th floor. Of course, the lift can't go up high than N,and can't go down lower than 1. For example, there is a buliding with 5 floors, and k1 = 3, k2 = 3,k3 = 1,k4 = 2, k5 = 5.Begining from the 1 st floor,you can press the button "UP", and you'll go up to the 4 th floor,and if you press the button "DOWN", the lift can't do it, because it can't go down to the -2 th floor,as you know ,the -2 th floor isn't exist.

Here comes the problem: when you are on floor A,and you want to go to floor B,how many times at least he has to press the button "UP" or "DOWN"?

Input

The input consists of several test cases.,Each test case contains two lines.

The first line contains three integers N ,A,B( 1 <= N,A,B <= 200) which describe above,

The second line consist N integers k1,k2,....kn.

A single 0 indicate the end of the input.

Output

For each case of the input output a interger, the least times you have to press the button when you on floor A,and you want to go to floor B.If you can't reach floor B,printf "-1".

Sample Input

5 1 53 3 1 2 50

Sample Output

3 题意： 有一天桐桐做了一个梦，梦见了一种很奇怪的电梯。  大楼的每一层楼都可以停电梯，而且第i层楼(1≤i≤N)上有一个数字K；(0≤Ki≤N)。  电梯只有四 个按钮：开，关，上，下。上下的层数等于当前楼层上的那个数字。当然，如果不能满足要求，相应的按钮就会失灵。  例如：3 3 1 2 5代表了Ki (K1=3，K2=3，…)，从一楼开始。在一楼，按 “上” 可以到4楼，按 “下” 是不起作用的，因为没有-2楼。  那么，从A楼到B楼至少要按几次按钮 呢？ 思路：bfs,见下方代码详细注释

//// Created by hjy on 2019/7/10.//#include
      
       #include
       
        #include
        
         #include
         
          //#include
          
           using namespace std;typedef long long ll;struct node{    int x;    int dept;};//结构体存储位置和步数int A,B;//起点和终点int book[205];//记录是否走过int put[205];//存储题给数组数值int n;//多少层int bfs(){    queue
           
            qu;//以结构体压入队列中 node now,next;//建立现在状态和下一步状态 now.x=A; now.dept=0;//现在状态的起始位置是A，步数为0； book[A]=1;//起始位置走过 qu.push(now);//将起始位置压入队列中 while(!qu.empty()) { now=qu.front();//现在状态为队首的状态 qu.pop();//删除队首，进行下一步操作 if(now.x==B)//如果走到了B终点状态，返回步数值 return now.dept; for(int i=-1;i<=1;i+=2)//小技巧，上下两个方向，往下走乘以-1，上走乘以1 { next=now;//下一步状态先令他为现在状态 next.x+=(put[next.x]*i);//下一步状态为现在状态加上1乘以或-1乘以原数组该位置的值 if(next.x<=0||next.x>n||book[next.x])//判断是否越界 continue; book[next.x]=1;//下位置就走过了 next.dept++;//步数加一 qu.push(next);//将下一位置压入队列中 } } return -1;//否则返回-1；}int main(){ while(cin>>n&&n) { cin>>A>>B; for(int i=1;i<=n;i++) cin>>put[i]; memset(book,0,sizeof(book));//初始化数组 cout<
            
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