B

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (0 ≤N ≤ 100,000) on a number line and the cow is at a pointK (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 orX+ 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 andK

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.

本题思路：

哎我是copy的

主要在于利用队列 这个广搜

c++代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int maxn=100009;
int book[maxn]={0};
int step[maxn]={0};
queue<int>q;
int bfs(int n,int k)
{
    int i;
    int t;
    book[n]=1;
    step[n]=0;
    q.push(n);
    while(q.empty()==0)
    {
        int head=q.front();
        q.pop();
        for(i=1;i<=3;i++)
        {
            if(i==1)
            t=head+1;
            else if(i==2)
            t=head-1;
            else if(i==3)
            t=head*2;
            if(t<0||t>maxn)
            continue;
            if(book[t]==0)
            {
            q.push(t);
            book[t]=1;
            step[t]=step[head]+1;
            }
            if(t==k)    
            return step[t];
        }
    }
}
int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        memset(book,0,sizeof(book));
           memset(step,0,sizeof(step));
        while(q.empty()==0)q.pop();
        if(n>=k)printf("%d\n",n-k);
        else if(n<k)printf("%d\n",bfs(n,k));
    }
    return 0;
}