zhber 有好多做过的题没写下来，如果我还能记得就补吧

## 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?

农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上幽发，尽快把那只奶牛抓回来．
他们都站在数轴上．约翰在N(O≤N≤100000)处，奶牛在K(O≤K≤100000)处．约翰有

那么，约翰需要多少时间抓住那只牛呢？

## Input

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

仅有两个整数N和K.

## Output

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

最短的时间．

## Sample Input

5 17
Farmer John starts at point 5 and the fugitive cow is at point 17.

## Sample Output

4

OUTPUT DETAILS:

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.

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int dist[500001];
int q[500001];
inline bool mark(int x)
{
return !(x<0||x>max(2*m+1,n+1));
}
int main()
{
freopen("catchcow.in","r",stdin);
freopen("catchcow.out","w",stdout);
scanf("%d%d",&n,&m);
memset(dist,-1,sizeof(dist));
q[1]=n;dist[n]=0;
{
if(mark(now)&&dist[now]==-1)
{
q[++tail]=now;
}
if(mark(now)&&dist[now]==-1)
{
q[++tail]=now;
}