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

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 <iostream>

#include <stdio.h>

#include <queue>

#include <cstring>

using namespace std;

const int Max = ;

int s, e;

struct node

{

    int x, step;

}now, Next;

bool v[];

queue<node>q;

int bfs()

{

    now.x = s;

    now.step = ;

    v[s] = ;

    q.push(now);

    while (!q.empty()) {

        now = q.front();

        q.pop();

        if (now.x == e)

            return now.step;

        Next.x = now.x*;

        Next.step = now.step+;

        if (Next.x >=  && Next.x <= Max && !v[Next.x]){

            q.push(Next);

            v[Next.x] = ;

        }

        Next.x = now.x-;

        Next.step = now.step+;

        if (Next.x >=  && Next.x <= Max && !v[Next.x]){

            q.push(Next);  //若先检查v[Next.x] 会出现数组越界v[-1]的情况

            v[Next.x] = ;

        }

        Next.x = now.x+;

        Next.step = now.step+;

        if (Next.x >=  && Next.x <= Max && !v[Next.x]){

            q.push(Next);

            v[Next.x] = ;

        }

    }

    return ;

}

int main()

{

    //freopen("1.txt", "r", stdin);

    cin >> s >> e;

    memset(v, , sizeof(v));

    cout << bfs();

    return ;

}