麻豆小视频在线观看_中文黄色一级片_久久久成人精品_成片免费观看视频大全_午夜精品久久久久久久99热浪潮_成人一区二区三区四区

首頁 > 學院 > 開發(fā)設計 > 正文

【Codeforces 735 C Tennis Championship】+ 貪心

2019-11-11 06:42:00
字體:
供稿:網(wǎng)友

C. Tennis Championship time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

Famous Brazil city Rio de Janeiro holds a tennis tournament and Ostap Bender doesn’t want to miss this event. There will be n players participating, and the tournament will follow knockout rules from the very first game. That means, that if someone loses a game he leaves the tournament immediately.

Organizers are still arranging tournament grid (i.e. the order games will happen and who is going to play with whom) but they have already fixed one rule: two players can play against each other only if the number of games one of them has already played differs by no more than one from the number of games the other one has already played. Of course, both players had to win all their games in order to continue participating in the tournament.

Tournament hasn’t started yet so the audience is a bit bored. Ostap decided to find out what is the maximum number of games the winner of the tournament can take part in (assuming the rule above is used). However, it is unlikely he can deal with this PRoblem without your help. Input

The only line of the input contains a single integer n (2?≤?n?≤?1018) — the number of players to participate in the tournament. Output

Print the maximum number of games in which the winner of the tournament can take part. Examples Input

2

Output

1

Input

3

Output

2

Input

4

Output

2

Input

10

Output

4

Note

In all samples we consider that player number 1 is the winner.

In the first sample, there would be only one game so the answer is 1.

In the second sample, player 1 can consequently beat players 2 and 3.

In the third sample, player 1 can’t play with each other player as after he plays with players 2 and 3 he can’t play against player 4, as he has 0 games played, while player 1 already played 2. Thus, the answer is 2 and to achieve we make pairs (1,?2) and (3,?4) and then clash the winners.

得分為n的人與得分為n - 1的比賽才能使比賽次數(shù)最大化,多少人才能出一個得分為N的人呢?

AC代碼:

#include<cstdio>typedef long long LL;int main(){ LL N,l = 1, r = 1,num = 0,t; scanf("%lld",&N); while(l + r <= N) t = l,l = r,r = t + r,num++; printf("%lld/n",num); return 0;}
發(fā)表評論 共有條評論
用戶名: 密碼:
驗證碼: 匿名發(fā)表
主站蜘蛛池模板: 免费中文视频 | 国产做爰全免费的视频黑人 | 免费观看视频91 | 日本在线视频一区二区三区 | 久久蜜桃香蕉精品一区二区三区 | 黄色片网站在线免费观看 | 日本不卡二区 | 国产寡妇xxxxxxxx性开放 | 亚洲天堂成人在线 | 中文日韩 | 久久色播 | 日韩视频不卡 | 一级黄色毛片免费 | 亚洲人成网站免费播放 | 欧美性成人| 日本网站在线看 | 欧美成年性h版影视中文字幕 | 免费观看一级 | 亚洲一区 国产 | 最新黄色电影网站 | 日本最新免费二区三区 | 日朝毛片 | 31freehdxxxx欧美| 亚洲国产精品二区 | 精品国产一级毛片 | 成人精品一区二区三区中文字幕 | 成人在线观看污 | 国产资源在线视频 | 一级成人在线 | 嗯~啊~弄嗯~啊h高潮视频 | 中文日韩 | 久久精品亚洲国产奇米99 | 亚洲极色| 亚洲国产精品500在线观看 | 黄色影院在线观看视频 | www视频免费在线观看 | 亚洲精品成人18久久久久 | 黄色影院在线看 | 亚洲码无人客一区二区三区 | 国产精品国产成人国产三级 | 日韩美女电影 |