hdu5802 Windows 10(贪心搜索)
2016-08-04 21:59
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思路:30w个样例1s...完全没有往搜索那个方面去想,一直想找一下规律...智障啊...
题解:
_您可能是正版Windows 10的受害者_ 直接贪心就好
比较直观的看法是使劲往下降,然后升回来
或者使劲往下降然后停顿然后再使劲往下降。。。
于是就能将问题变成一个子问题,然后dfs就好
需要注意的是由于按up键也可以打断连续向下的功效
所以应该记录停顿了几次,以后向上的时候用停顿补回来
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
2
1 5
7 3
Sample Output
4
4
Author
UESTC
Source
2016 Multi-University Training Contest 6
题解:
_您可能是正版Windows 10的受害者_ 直接贪心就好
比较直观的看法是使劲往下降,然后升回来
或者使劲往下降然后停顿然后再使劲往下降。。。
于是就能将问题变成一个子问题,然后dfs就好
需要注意的是由于按up键也可以打断连续向下的功效
所以应该记录停顿了几次,以后向上的时候用停顿补回来
#include<bits/stdc++.h> using namespace std; #define LL long long LL p,q; LL dfs(LL pp,LL qq,LL step,LL stop) { if(qq==pp) return step; LL move = 0; while(pp-(1<<move)+1 > qq) move++; if(pp-(1<<move)+1==qq) return step+move; LL up = qq-max((LL)0,pp-(1<<move)+1); LL res = move+max((LL)0,up-stop); return min(step+res,dfs(pp-(1<<(move-1))+1,qq,step+move,stop+1)); } int main() { int T; scanf("%d",&T); while(T--) { scanf("%lld%lld",&p,&q); if(q>=p) { printf("%lld\n",q-p); continue; } printf("%lld\n",dfs(p,q,0,0)); } }
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
2
1 5
7 3
Sample Output
4
4
Author
UESTC
Source
2016 Multi-University Training Contest 6
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