1253: catch the cow

 

Description [title]

The farmer knows the location of a cow, you want to grab it. Cattle farmers and are located on the number line, the farmer starting at point N (0 ≦ N ≤100000)

Cattle at point K (0≤ K ≤100000)

. There are two ways to move the farmer:

1, from X

Moving the X- -1 or X- + 1'd

, Take a minute each move

2, to move from X 2 × X

, Take a minute each move

Assuming that the cattle farmers are not aware of the action, stand still. The farmer takes the least amount of time to seize the cattle?

[Enter]

Two integers, N

And K

。

[Output]

An integer cattle farmer caught a minimum number of minutes you want to spend.

[Sample input]

5 17

[Sample Output]

4

// Created on 2020/2/8

/*#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <climits>*/
#include <bits/stdc++.h>
#define MAXX 40*40+5

using namespace std;

//int i,j,k;

const int maxn=INT_MAX;

const int idata=100000+5;
//int aim[idata];
bool judge[idata];
bool flag;
int soux,souy,
    exitx,exity;
//int maps[idata][idata];
//const int change[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
long long n,m;
long long Count[idata];
long long nx,ny;
long long x[idata];

int main()
{
    long long head,tail;
    int i;
    cin>>n>>m;

    head=0,tail=1;
    x[1]=n;
    judge[n]=true;
    Count[n]=0;

    while(head<tail)
    {
        head++;

        for(i=0;i<3;i++)
        {
            switch(i)
            {
                case 0:nx=x[head]+1;break;
                case 1:nx=x[head]-1;break;
                case 2:nx=x[head]*2;break;
            }

            if(nx<=idata&&nx>=0&&!judge[nx])
            {
                judge[nx]=true;
                tail++;
                Count[nx]=Count[x[head]]+1;
                x[tail]=nx;
            }

            if(nx==m)
            {
                flag=true;
                break;
            }
        }
        if(flag) break;
    }
    cout<<Count[m]<<endl;
    return 0;
}