hdu 5802 Windows 10 思维

Long long ago, there was an old monk living on the top of a mountain. Recently, our old monk found the operating system of his computer was updating to windows 10 automatically and he even can't just stop it !! 
With a peaceful heart, the old monk gradually accepted this reality because his favorite comic LoveLive doesn't depend on the OS. Today, like the past day, he opens bilibili and wants to watch it again. But he observes that the voice of his computer can be represented as dB and always be integer. 
Because he is old, he always needs 1 second to press a button. He found that if he wants to take up the voice, he only can add 1 dB in each second by pressing the up button. But when he wants to take down the voice, he can press the down button, and if the last second he presses the down button and the voice decrease x dB, then in this second, it will decrease 2 * x dB. But if the last second he chooses to have a rest or press the up button, in this second he can only decrease the voice by 1 dB. 
Now, he wonders the minimal seconds he should take to adjust the voice from p dB to q dB. Please be careful, because of some strange reasons, the voice of his computer can larger than any dB but can't be less than 0 dB.

Input

First line contains a number TT (1≤T≤3000001≤T≤300000),cases number. 
Next T line,each line contains two numbers pp and qq (0≤p,q≤109)(0≤p,q≤109)

Output

The minimal seconds he should take

Sample Input

21 57 3

Sample Output

4
4
题意：有3种操作 1：向下 连续向下的话，这次向下是 上次的*2
2：休息，不动 下次向下从 1开始
3：往上走一步
贪心的想法是，走到离q最近的上(1<<k)-1 和 下（1<<(k+1))-1，我们贪心的觉得这样走到它下面不断往上走，和在上界休息+1然后再往下是最优的，最后往上走的时候减去休息的时间就是最小的时间了，因为休息和往上走一步的时间是一样的
#include <bits/stdc++.h>using namespace std;typedef long long ll;ll a,b;ll log(ll x,ll y){ll k=0;while((1ll<<k)-1ll<(x-y)) k++;return k;}ll dfs(ll  x,ll y,ll step,int rest){if(x==y) return step;ll k=log(x,y);if(max(0ll,x-(1ll<<k)+1ll)==y) return step+k;ll up=y-max(0ll,x-(1ll<<k)+1ll);ll tmp=max(0ll,up-rest);return min(tmp+k+step,dfs(x-(1ll<<(k-1))+1ll,y,step+k,rest+1));}int main(){int t;scanf("%d",&t);while(t--){scanf("%lld%lld",&a,&b);if(a<=b) printf("%lld\n",b-a );else printf("%lld\n",dfs(a,b,0,0) );}}