Codeforces Round #382 (Div. 2) C. Tennis Championship 斐波那契
2 seconds
256 megabytes
standard input
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.
The only line of the input contains a single integer n (2 ≤ n ≤ 1018) — the number of players to participate in the tournament.
Print the maximum number of games in which the winner of the tournament can take part.
2
1
3
2
4
2
10
4
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.
题意:给你n个人,一个人有一个比赛次数,输了就淘汰,每个人能跟比赛次数相差不超过1的比赛,问一个赢的人最多比几次;
思路:显然4需要跟3比才能得到5,相当于n-1需要跟n-2比得到n,不就是一个菲薄那契么,暴力求解就好了;
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define esp 0.00000000001
const int N=1e5+,M=1e6+,inf=1e9;
const ll INF=1e18+;
ll a[];
int main()
{
a[]=;
a[]=;
for(int i=;i<=;i++)
a[i]=a[i-]+a[i-];
ll x;
scanf("%lld",&x);
for(int i=;i>=;i--)
if(x>=a[i])
{
printf("%d\n",i);
return ;
}
return ;
}
Codeforces Round #382 (Div. 2) C. Tennis Championship 斐波那契的更多相关文章
- Codeforces Round #382 (Div. 2) C. Tennis Championship
C. Tennis Championship time limit per test 2 seconds memory limit per test 256 megabytes input stand ...
- Codeforces Round #382 (Div. 2)C. Tennis Championship 动态规划
C. Tennis Championship 题目链接 http://codeforces.com/contest/735/problem/C 题面 Famous Brazil city Rio de ...
- codeforces Codeforces Round #597 (Div. 2) Constanze's Machine 斐波拉契数列的应用
#include<bits/stdc++.h> using namespace std; ]; ]; ; int main() { dp[] = ; scanf(); ); ; i< ...
- Codeforces Round #382 Div. 2【数论】
C. Tennis Championship(递推,斐波那契) 题意:n个人比赛,淘汰制,要求进行比赛双方的胜场数之差小于等于1.问冠军最多能打多少场比赛.题解:因为n太大,感觉是个构造.写写小数据, ...
- Codeforces Round #382 (Div. 2) (模拟|数学)
题目链接: A:Ostap and Grasshopper B:Urbanization C:Tennis Championship D:Taxes 分析:这场第一二题模拟,三四题数学题 A. 直接模 ...
- Codeforces Round #382 (Div. 2) 继续python作死 含树形DP
A - Ostap and Grasshopper zz题能不能跳到 每次只能跳K步 不能跳到# 问能不能T-G 随便跳跳就可以了 第一次居然跳越界0.0 傻子哦 WA1 n,k = map ...
- Codeforces Round #382 (Div. 2) D. Taxes 哥德巴赫猜想
D. Taxes 题目链接 http://codeforces.com/contest/735/problem/D 题面 Mr. Funt now lives in a country with a ...
- Codeforces Round #382 (Div. 2)B. Urbanization 贪心
B. Urbanization 题目链接 http://codeforces.com/contest/735/problem/B 题面 Local authorities have heard a l ...
- Codeforces Round #283 (Div. 2) D. Tennis Game(模拟)
D. Tennis Game time limit per test 2 seconds memory limit per test 256 megabytes input standard inpu ...
随机推荐
- IT职业选择与定位
(一) 位置有很多,最适合你的是哪个? 有的人在电子技术的层面工作,开发出性能强劲的芯片和硬件产品:有的人在别人开发的芯片和硬件上开发各种操作系统和驱动程序:有的人在各种操作系统或设备 ...
- bootstrap实现 手机端滑动效果,滑动到下一页,jgestures.js插件
bootstrap能否实现 手机端滑动效果,滑动到下一页 jgestures.js插件可以解决,只需要引入一个JS文件<script src="js/jgestures.min.js& ...
- TM1680的I2C的51例程
搞到一个例程,虽然是51的, 但是我的ST版本也是用的模拟I2C, 分析一下吧: unsigned char i=0;TM1680start(); //I2C起始信号 TM1680SendByte( ...
- makefile 中 $@ $^ %< 使用【转】
转自:http://blog.csdn.net/kesaihao862/article/details/7332528 这篇文章介绍在LINUX下进行C语言编程所需要的基础知识.在这篇文章当中,我们将 ...
- Temporary TempDB Tables [AX 2012]
Temporary TempDB Tables [AX 2012] 1 out of 4 rated this helpful - Rate this topic Updated: November ...
- jQuery添加删除元素
$(document).ready(function () { $('#radioExtranet').on('click', function () { showProjectInformation ...
- App store 如何使用 promo code | app store 打不开精品推荐和排行榜
1. app store 如何使用 promo code: 在app store的 右下角精品推荐标签页,拉到最下面 点击“兑换” ,跳转到新的页面,输入兑换码,然后右上角“兑换”,程序开始自动下载并 ...
- 20145227 《Java程序设计》第3周学习总结
20145227 <Java程序设计>第3周学习总结 教材学习内容总结 第四章 认识对象 4.1 类与对象 1.定义类:生活中描述事物无非就是描述事物的属性和行为.如:人有身高,体重等属性 ...
- springMVC配置freemarker 二(问题讨论篇)
上面一篇我已经说明了如何去配置freemarker,这里我就谈谈遇到的问题吧. 首先, 为什么要删除上面之前的.你要使用freemarkerviewresolver和上面的冲突了,因此要注释掉上面的. ...
- SQL Server 索引和表体系结构(一)
转自:http://www.cnblogs.com/chenmh/p/3780221.html 聚集索引 概述 关于索引和表体系结构的概念一直都是讨论比较多的话题,其中表的各种存储形式是讨论的重点,在 ...