Windows 10

Problem Description

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 T (1≤T≤300000),cases number.
Next T line,each line contains two numbers p and q (0≤p,q≤109)

 

Output

The minimal seconds he should take

 

Sample Input

 

Sample Output

 

Author

UESTC

 

Source

2016 Multi-University Training Contest 6

/**
题目：hdu5802 Windows 10
链接：http://acm.hdu.edu.cn/showproblem.php?pid=5802
题意：调节音量p到q，上升音量每秒只能上升1，下降音量每秒为2*x，x为上一次下降的音量，如果下降时休息或者上升音量则x置为1，音量最低为0
题意有问题：题目说p不能降为负数。但是如果p-x<0；那么可以把p-x<0的结果都看做p-x==0；

思路：贪心的去选，每次下降到终点上的最近一点，或者终点下的一点，取得一个最小值就好。

*/

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
const int mod=1e9+7;
const int maxn=1e6+5;
const double eps = 1e-12;
LL dfs(LL p,LL q,LL delay)
{
    LL x = 1;
    LL cnt = 0;
    while(p-x>=q){
        cnt++;
        p = p-x;
        x = x*2;
    }
    if(p==q) return cnt;
    return min(max(0LL,q-max(0LL,(p-x))-delay)+cnt+1,dfs(p,q,delay+1)+cnt+1);

}
int main()
{
    int T;
    LL p, q;
    cin>>T;
    while(T--)
    {
        scanf("%lld%lld",&p,&q);
        if(p<q){
            printf("%lld\n",q-p);
        }else
        {
            printf("%lld\n",dfs(p,q,0));
        }
    }

    return 0;
}