1.1013

2.

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 <= 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 5 3 3 1 2 5 0  Sample Output 3

3.電梯每層有一個數字，每個數字不相同，若第n層有個數字k，那麼這一層只能上k層或下k層，當然了不能低於1層或高於n層，給定起點與終點，求出最少要按幾次鍵

4.這道題用搜索的思想可以做出，廣度搜索

5.#include<iostream>
#include<string.h>
#include<set>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<numeric>
#include<math.h>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<queue>
#include<iomanip>
#include<cstdio>
using namespace std;
struct node
{
   int x ;
   int t ;
}n1,n2,m;
int c[2000],ww[2000];  
int flag ;                 
int n , a , b ;        
int main(  )
{
    while( scanf("%d" , &n ) , n )
    {
       if( n== 0 ) break ;      
       cin>>a>>b;
       for( int i= 1;  i<= n ; i++ )     
       { cin>>ww[i] ; c[i] = 0 ; }
       flag = 0 ;                      //初始為0
       n1.x = a ; n1.t = 0 ;        //把a開始所在樓層賦給結構體， 所走步數為0
       queue<node>Q;              
       Q.push(n1);                        
       c[n1.x] = 1 ;          
       while( !Q.empty() )       
       {
          m = Q.front() ;          
          Q.pop();                  
          if( m.x == b )         
          { flag = 1 ; break ; }
          n1.x = m.x - ww[m.x];  
          n2.x = m.x + ww[m.x]; 
          if( n1.x>0 && n1.x<= b && !c[n1.x] ）
          {
                n1.t = m.t + 1 ;       
                c[n1.x] = 1 ;     
                Q.push(n1);
              }
            if( n2.x>0 && n2.x<= b && !c[n2.x] )
             {
                n2.t = m.t+1 ;
                c[n2.x] = 1 ;
                Q.push(n2);
             }
       }
       if( flag ) cout<<m.t<<endl;
       else cout<<"-1"<<endl;
    }
    return 0;
}