<span style="color:#330099;">/*

A - 广搜 基础

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit

Status

Description

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?

Input

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

Output

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

Sample Input

5 17

Sample Output

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.

By Grant Yuan

2014.7.13

*/

#include<iostream>

#include<stdio.h>

#include<string.h>

#include<queue>

#include<cstdio>

using namespace std;

bool a[100003];

typedef struct{

   int num;

   int sum;

}dd;

queue<dd>q;

int n,k;

int res;

void slove()

{int m,count;

   int x;

   dd init;

   while(!q.empty()){

       if(q.front().num==n)

       {  res=q.front().sum;

           break;

       }

    m=q.front().num;

    count=q.front().sum;

   x=m-1;

   if(a[x]==0)

     {   init.num=x;

         init.sum=count+1;

         q.push(init);

         a[x]=1;

     }

    x=m+1;

     if(a[x]==0)

     {

        init.num=x;

         init.sum=count+1;

         q.push(init);

         a[x]=1;

     }

     if(m%2==0){

         x=m/2;

      if(a[x]==0)

     {

         init.num=x;

         init.sum=count+1;

         q.push(init);

         a[x]=1;

     }}

     q.pop();}

}

int main()

{

       scanf("%d%d",&n,&k);

    memset(a,0,sizeof(a));

    dd init;

    init.num=k;

    init.sum=0;

    a[k]=1;

    q.push(init);

    slove();

    cout<<res<<endl;

    return 0;

}

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