如何确定的一轮元素树（比赛括号内）？

本文介绍了如何确定的一轮元素树（比赛括号内）？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

假设我们有以下三种：

1     9 2         13 3     10 4             15 五     11 6         14 7     12 8

元素在哪里（匹配）： 1-8是圆的1 9-12是圆的2 13-14是第3轮 15轮4

我怎么能确定的在shuch树轮的元素N？

我目前的公式是：

total_rounds =地板（日志（totalTeams，2））; matches_per_round =（totalTeams / POW（2 current_round）） next_match_id =（totalTeams / 2）+ CEIL（match_id / 2） total_matches = total_teams - 1

解决方案

想象一下，树也被列在相反的。

15      7 14          3 13      6 12              1 11      五 10          2 9      4 8

在这种情况下，它会简单地将数的对数，四舍五入。现在，我们只需要减去此数轮数，我们就大功告成了。

reverse_number = total_matches - match_number + 1; reverse_match_round =地板（日志（reverse_number，2））; match_round = total_rounds - match_round;

（注意， reverse_match_round 实际上是0索引，不像match_round。但是，由于我们从 total_rounds 减去它，它更容易保持这种方式比1索引它。如果你preFER这1索引，只需添加 +1 到每个最后两行。）

Assume we have following tree:

1 9 2 13 3 10 4 15 5 11 6 14 7 12 8

Where elements(matches): 1-8 is round 1 9-12 is round 2 13-14 is round 3 15 is round 4

How I can determinate round of element "n" in shuch tree?

My current formulas are:

total_rounds = floor(log(totalTeams,2)); matches_per_round = (totalTeams / pow(2, current_round)) next_match_id = (totalTeams/2) + ceil(match_id/2) total_matches = total_teams - 1

解决方案

Imagine the tree was numbered in reverse.

15 7 14 3 13 6 12 1 11 5 10 2 9 4 8

In that case, it'd simply be the logarithm of the number, rounded down. Now we simply subtract this number from the number of rounds, and we're done.

reverse_number = total_matches - match_number + 1; reverse_match_round = floor(log(reverse_number, 2)); match_round = total_rounds - match_round;

(Note, reverse_match_round is actually 0-indexed, unlike match_round. However, since we subtract it from total_rounds, it's easier to keep it that way than to 1-index it. If you prefer it 1-indexed, simply add +1 to each of the last two lines.)