Farmer John has been informed of the location of a fugitive（逃亡的;难以捉摸的;短暂的) 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 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 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(追赶;工作), does not move at all, how long does it take for Farmer John to retrieve it?

#include<stdio.h>#include<string.h>int dx[3]={-1, 1, 2};int vis[3000000];int n, k;struct node{    int x, ans;}q[3000000], t, f;void bfs(){    memset(vis, 0, sizeof(vis));    memset(q, 0, sizeof(q));    int s=0, e=0;    vis[n] = 1;    q[s++].x = n;    t.ans = 0;    while(s>e)    {        t = q[e++];        for(int i=0; i<3; i++)        {            if(i!=2)                f.x = t.x+dx[i];            else                f.x = t.x*dx[i];            f.ans = t.ans+1;            if(f.x==k)            {                printf("%d\n", f.ans);                return;            }            else{                if(!vis[f.x] && (f.x>0 && f.x<100000) )                {                       vis[f.x] = 1;                    q[s++] = f;                }            }        }    }}int main(){    while(~scanf("%d%d", &n, &k))    {        if(n>=k)        {            printf("%d\n", n-k);            continue;        }        else        {            bfs();        }    }    return 0;}