Catch That Cow

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting. * Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute * Teleporting: FJ can move from any point X to the point 2 × X in a single minute. If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Line 1: Two space-separated integers: N and K

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

5 17

4   
       
         
     Hint    
    The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

/*终于AC了  *///看了别人的代码 其实可以用结构体  建个数组记忆是否进过队  并且N不用那么大//N=100000+10就够了 因为太大也不可能是答案//优化空间可以改成  N=100000+10   //（不知道每次查长度需要时间不） 要的话改成结构体可以省时间 #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;const int N=100000*6+10;queue
         
           q;bool book[N];int main(){ int a,b,re,tmp,i; while(scanf("%d%d",&a,&b)==2){ while(!q.empty()) q.pop(); re=0; memset(book,0,sizeof(book)); book[a]=1; q.push(a); while(!q.empty()){ int len=q.size(); for(i=0;i
          
           0&&!book[tmp-1]){ book[tmp-1]=1; q.push(tmp-1); } if(tmp*2
           
            #include 
            
             #include 
             
              using namespace std;const int N=100000;struct Node{ int v,t;}node,tmp;queue
              
                q;int main(){ int a,b; while(scanf("%d%d",&a,&b)==2){ while(!q.empty()) q.pop(); node.v=a; node.t=0; q.push(node); while(!q.empty()){ node=q.front(); q.pop(); if(node.v==b) break; tmp.v=node.v+1; tmp.t=node.t+1; q.push(tmp); tmp.v=node.v-1; tmp.t=node.t+1; q.push(tmp); tmp.v=node.v*2; tmp.t=node.t+1; q.push(tmp); } printf("%d\n",node.t); } return 0;}*//*想水一下 水不过 TLE#include 
               
                #include 
                
                 using namespace std;const int N=100000;int b,re;void dfs(int now,int sum){ if(sum>re) return; else if(now==b){ if(sum