1014 A strange lift

Problem Description

There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 &lt;= Ki &lt;= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button &quot;UP&quot; , 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 &quot;DOWN&quot; , 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 &quot;UP&quot;, and you'll go up to the 4 th floor,and if you press the button &quot;DOWN&quot;, 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.<br>Here comes the problem: when you is on floor A,and you want to go to floor B,how many times at least he havt to press the button &quot;UP&quot; or &quot;DOWN&quot;?<br>

 

Input

The input consists of several test cases.,Each test case contains two lines.<br>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.<br>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 5<br>3 3 1 2 5<br>0

 

Sample Output

3

1. #include <cstdio>  
2. #include <cstdlib>  
3. #include <cstring>  
4. #include <cmath>  
5. #include <queue>  
6. #include <climits>  
7.   
8. using namespace std;  
9.   
10. const int MAX = 202;  
11. int flr[MAX],ans;  
12. int dist[MAX];  
13. int n,b;  
14.   
15. void dfs(int s,int cnt){  
16.     if(s>n || s<1)return;  
17.     if(s==b){  
18.         if(cnt<ans){  
19.             ans = cnt;  
20.             return;  
21.         }  
22.     }  
23.     if(cnt>=dist[s])return;  
24.     dist[s] = cnt;  
25.     dfs(s+flr[s],cnt+1);  
26.     dfs(s-flr[s],cnt+1);  
27. }  
28.   
29. int main(){  
30.     //freopen("in.txt","r",stdin);  
31.   
32.     int a,i,cnt;  
33.     while(scanf("%d",&n)!=EOF && n){  
34.         scanf("%d %d",&a,&b);  
35.         for(i=1;i<=n;++i){  
36.             scanf("%d",&flr[i]);  
37.             dist[i] = INT_MAX - 10;  
38.         }  
39.         cnt = 0;  
40.         ans = INT_MAX;  
41.         dfs(a,cnt);  
42.         if(ans==INT_MAX){  
43.             printf("-1\n");  
44.         }else{  
45.             printf("%d\n",ans);  
46.         }  
47.     }  
48.   
49.     return 0;  
50. }