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.

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

  1. 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 ...

  2. 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 ...

  3. 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< ...

  4. 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, ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. python 提取图片转为16 24BPP 的方法

    python 中处理图片用的是 pil ,在 linux  和 win 上都可以使用. centOS 5.x 上安装的方法是 yum install python-imaging 24BPP: imp ...

  2. 160826、浏览器渲染页面过程描述,DOM编程技巧以及重排和重绘

    一.浏览器渲染页过程描述   1.浏览器解析html源码,然后创建一个DOM树. 在DOM树中,每一个HTML标签都有一个对应的节点(元素节点),并且每一个文本也都有一个对应的节点(文本节点). DO ...

  3. webpack笔记_(2)_Refusing to install webpack as a dependency of itself

    安装webpack时,出现以下问题: Refusing to install webpack as a dependency of itself npm ERR! Windows_NT npm ERR ...

  4. PHP array_column() 函数

    定义和用法 array_column() 返回输入数组中某个单一列的值. array_column(array,column_key,index_key); 参数 描述 array 必需.规定要使用的 ...

  5. linux下对sh文件的操作

    1.创建test.sh文件 touch test.sh 2.编辑sh文件 vi test.sh(i:插入 | esc:退出insert模式 | wq+回车:退出) 3.保存退出 敲击esc, 然后输入 ...

  6. poj-3259-wormholes-spfa-判负环

    题意:N个顶点, M条双向边, W条权值为负的单向边.求是否存在负环. 思路:首先你要懂bellman-ford或spfa..这是基础的spfa判断是否存在负环的题,存在负环的节点会重复入队(因为最短 ...

  7. Javascript中自动切换焦点

      <!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title&g ...

  8. c#播放器

    using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; usin ...

  9. Linux 多线程应用中如何编写安全的信号处理函数

    http://blog.163.com/he_junwei/blog/static/1979376462014021105242552/ http://www.ibm.com/developerwor ...

  10. mongoDB Replica集群配置(1主+1从+1仲裁)

    1.mongoDB节点介绍 主节点(Primary) 在复制集中,主节点是唯一能够接收写请求的节点.MongoDB在主节点进行写操作,并将这些操作记录到主节点的oplog中.而从节点将会从oplog复 ...