HMMT 十一月 2010 · GEN2 赛 · 第 2 题

HMMT November 2010 — GEN2 Round — Problem 2

[ 4 ] 16 progamers are playing in another single elimination tournament. Each round, each of the remaining progamers plays against another and the loser is eliminated. Additionally, each time a progamer wins, he will have a ceremony to celebrate. A player’s first ceremony is ten seconds long, and afterward each ceremony is ten seconds longer than the last. What is the total length in seconds of all the ceremonies over the entire tournament?

解析

[ 4 ] 16 progamers are playing in another single elimination tournament. Each round, each of the remaining progamers plays against another and the loser is eliminated. Additionally, each time a progamer wins, he will have a ceremony to celebrate. A player’s first ceremony is ten seconds long, and afterward each ceremony is ten seconds longer than the last. What is the total length in seconds of all the ceremonies over the entire tournament? Answer: 260 At the end of the first round, each of the 8 winners has a 10 second ceremony. After the second round, the 4 winners have a 20 second ceremony. The two remaining players have 30 second ceremonies after the third round, and the winner has a 40 second ceremony after the finals. So, all of the ceremonies combined take 8 · 10 + 4 · 20 + 2 · 30 + 40 = 260 seconds.