A. Tennis Tournament

题目连接:

http://www.codeforces.com/contest/628/problem/A

Description

A tennis tournament with n participants is running. The participants are playing by an olympic system, so the winners move on and the losers drop out.

The tournament takes place in the following way (below, m is the number of the participants of the current round):

let k be the maximal power of the number 2 such that k ≤ m,

k participants compete in the current round and a half of them passes to the next round, the other m - k participants pass to the next round directly,

when only one participant remains, the tournament finishes.

Each match requires b bottles of water for each participant and one bottle for the judge. Besides p towels are given to each participant for the whole tournament.

Find the number of bottles and towels needed for the tournament.

Note that it's a tennis tournament so in each match two participants compete (one of them will win and the other will lose).

Input

The only line contains three integers n, b, p (1 ≤ n, b, p ≤ 500) — the number of participants and the parameters described in the problem statement.

Output

Print two integers x and y — the number of bottles and towels need for the tournament.

Sample Input

5 2 3

Sample Output

20 15

Hint

题意

有两种水,n个人参加比赛

每次都会选择出小于等于n的最大2的倍数,然后让这些人比赛,每个参加比赛的人可以获得b瓶A水,裁判也得有一瓶A水

然后每个人都会获得p瓶B水

然后问你打完所有比赛后,需要多少瓶A水,多少瓶B水

题解:

A题就不要想太多,直接暴力吧……

虽然O(1)公式也有

代码

#include<bits/stdc++.h>
using namespace std; vector<int> two;
int main()
{
long long n,b,p;
cin>>n>>b>>p;
long long ans = 0,ans2 = n*p;
while(n>1)
{
int t = (n)/2*2;
ans+=t*b+t/2;
n-=t/2;
}
cout<<ans<<" "<<ans2<<endl;
}

Educational Codeforces Round 8 A. Tennis Tournament 暴力的更多相关文章

  1. Codeforces Educational Codeforces Round 8 A. Tennis Tournament

    大致题意: 网球比赛,n个參赛者,每场比赛每位选手b瓶水+裁判1瓶水,所有比赛每一个參赛者p条毛巾 每一轮比赛有2^k个人參加比赛(k为2^k<=n中k的最大值),下一轮晋级人数是本轮每场比赛的 ...

  2. Educational Codeforces Round 1 A. Tricky Sum 暴力

    A. Tricky Sum Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/598/problem ...

  3. Educational Codeforces Round 8 B. New Skateboard 暴力

    B. New Skateboard 题目连接: http://www.codeforces.com/contest/628/problem/A Description Max wants to buy ...

  4. Educational Codeforces Round 19 E. Array Queries(暴力)(DP)

    传送门 题意 给出n个数,q个询问,每个询问有两个数p,k,询问p+k+a[p]操作几次后超过n 分析 分块处理,在k<sqrt(n)时,用dp,大于sqrt(n)用暴力 trick 代码 #i ...

  5. Codeforces Educational Codeforces Round 17 Problem.A kth-divisor (暴力+stl)

    You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist ...

  6. Educational Codeforces Round 37

    Educational Codeforces Round 37 这场有点炸,题目比较水,但只做了3题QAQ.还是实力不够啊! 写下题解算了--(写的比较粗糙,细节或者bug可以私聊2333) A. W ...

  7. Educational Codeforces Round 35 (Rated for Div. 2)

    Educational Codeforces Round 35 (Rated for Div. 2) https://codeforces.com/contest/911 A 模拟 #include& ...

  8. Educational Codeforces Round 59 (Rated for Div. 2) DE题解

    Educational Codeforces Round 59 (Rated for Div. 2) D. Compression 题目链接:https://codeforces.com/contes ...

  9. Educational Codeforces Round 32

    http://codeforces.com/contest/888 A Local Extrema[水] [题意]:计算极值点个数 [分析]:除了第一个最后一个外,遇到极值点ans++,包括极大和极小 ...

随机推荐

  1. 通过or注入py脚本

    代码思路 1.主要还是参考了别人的代码,确实自己写的和别人写的出路很大,主要归咎还是自己代码能力待提高吧. 2.将功能集合成一个函数,然后通过*args这个小技巧去调用.函数的参数不是argv的值,但 ...

  2. 盲注脚本2.基于bool

    盲注脚本2.基于bool #!/usr/bin/env python #encoding:utf-8 #by i3ekr #using # python sqlinject.py -D "数 ...

  3. mysql之基本数据库操作(二)

    环境信息 数据库:mysql-5.7.20 操作系统:Ubuntu-16.04.3 mysql的启动.退出.重启 # 启动 $ sudo service mysqld start # 停止 $ sud ...

  4. 经典卷积网络模型 — LeNet模型笔记

    LeNet-5包含于输入层在内的8层深度卷积神经网络.其中卷积层可以使得原信号特征增强,并且降低噪音.而池化层利用图像相关性原理,对图像进行子采样,可以减少参数个数,减少模型的过拟合程度,同时也可以保 ...

  5. CSS原生布局方式

    前言 网页原生布局的方法其实网上有很多,大概为Flow(流动布局模型).Float(浮动布局模型).Layer(层级布局模型).<!--more--> Flow布局 流动布局模型其实就是默 ...

  6. [How to] 动态布局可变高度的cell的应用

    1.简介 代码:https://github.com/xufeng79x/DynamicChangeableCell 微博界面,微信和QQ聊天界面,这些界面的布局大都不确定,且每一条消息的高度也不一样 ...

  7. C#ActiveX控件开发

    1.新建项目,选择C#,选择.NET Framework2.0,新建一个Windows窗体控件库项目,命名为ActiveXDemo; 2.右击ActiveXDem项目,选择属性——应用程序——程序集信 ...

  8. 关于那些Android中不常用的设置属性

    很多在manifest中的属性我们经常遗忘了它们,或者经常看到但又不是很明白它的作用.那么在这里我就拿了一些属性简单的解释一下,防止以后碰到却不知道其中的意思.不是很全,以后会断断续续的补充吧 一.a ...

  9. Python初学--字符串

    ASCII.Unicode和UTF-8的关系 在计算机内存中,统一使用Unicode编码,当需要保存到硬盘或者需要传输的时候,就转换为UTF-8编码 记事本编辑的时候,从文件读取的UTF-8字符被转换 ...

  10. POJ 1177 Picture(线段树:扫描线求轮廓周长)

    题目链接:http://poj.org/problem?id=1177 题目大意:若干个矩形,求这些矩形重叠形成的图形的轮廓周长. 解题思路:这里引用一下大牛的思路:kuangbin 总体思路: 1. ...