B. Simple Game( Codeforces Round #316 (Div. 2) 简单题)
1 second
256 megabytes
standard input
standard output
One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from 1 to n.
Let's assume that Misha chose number m, and Andrew chose number a.
Then, by using a random generator they choose a random integer c in the range between 1 and n (any
integer from 1 to n is
chosen with the same probability), after which the winner is the player, whose number was closer to c. The boys agreed that if m and a are
located on the same distance from c, Misha wins.
Andrew wants to win very much, so he asks you to help him. You know the number selected by Misha, and number n. You need to determine
which value of a Andrew must choose, so that the probability of his victory is the highest possible.
More formally, you need to find such integer a (1 ≤ a ≤ n),
that the probability that is
maximal, where c is the equiprobably chosen integer from 1 to n (inclusive).
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 109)
— the range of numbers in the game, and the number selected by Misha respectively.
Print a single number — such value a, that probability that Andrew wins is the highest. If there are multiple such values, print the
minimum of them.
3 1
2
4 3
2
In the first sample test: Andrew wins if c is equal to 2 or 3.
The probability that Andrew wins is 2 / 3. If Andrew chooses a = 3,
the probability of winning will be 1 / 3. If a = 1,
the probability of winning is 0.
In the second sample test: Andrew wins if c is equal to 1 and 2.
The probability that Andrew wins is 1 / 2. For other choices of a the
probability of winning is less.
求选哪个点使得|c-a| < |c-m|的概率最高
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<stack>
#include<vector>
#include<map> using namespace std; int n,m; int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n == 1)
{
printf("1\n");
}
else if(2*m<=n)
{
printf("%d\n",m+1);
}
else
{
printf("%d\n",m-1);
}
}
return 0;
}
B. Simple Game( Codeforces Round #316 (Div. 2) 简单题)的更多相关文章
- B. Vanya and Books( Codeforces Round #308 (Div. 2) 简单题)
B. Vanya and Books time limit per test 1 second memory limit per test 256 megabytes input standard i ...
- B. Case of Fake Numbers( Codeforces Round #310 (Div. 2) 简单题)
B. Case of Fake Numbers time limit per test 2 seconds memory limit per test 256 megabytes input stan ...
- Codeforces Round #378 (Div. 2) D题(data structure)解题报告
题目地址 先简单的总结一下这次CF,前两道题非常的水,可是第一题又是因为自己想的不够周到而被Hack了一次(或许也应该感谢这个hack我的人,使我没有最后在赛后测试中WA).做到C题时看到题目情况非常 ...
- Codeforces Round #612 (Div. 2) 前四题题解
这场比赛的出题人挺有意思,全部magic成了青色. 还有题目中的图片特别有趣. 晚上没打,开virtual contest打的,就会前三道,我太菜了. 最后看着题解补了第四道. 比赛传送门 A. An ...
- Codeforces Round #552 (Div. 3) A题
题目网址:http://codeforces.com/contest/1154/problem/ 题目意思:就是给你四个数,这四个数是a+b,a+c,b+c,a+b+c,次序未知要反求出a,b,c,d ...
- Codeforces Round #713 (Div. 3)AB题
Codeforces Round #713 (Div. 3) Editorial 记录一下自己写的前二题本人比较菜 A. Spy Detected! You are given an array a ...
- Codeforces Round #412 Div. 2 补题 D. Dynamic Problem Scoring
D. Dynamic Problem Scoring time limit per test 2 seconds memory limit per test 256 megabytes input s ...
- Codeforces Round #316 (Div. 2) B. Simple Game
思路:把n分成[1,n/2],[n/2+1,n],假设m在左区间.a=m+1,假设m在右区间,a=m-1.可是我居然忘了处理1,1这个特殊数据.被人hack了. 总结:下次一定要注意了,提交前一定要看 ...
- Codeforces Round #316 (Div. 2B) 570B Simple Game 贪心
题目:Click here #include <bits/stdc++.h> using namespace std; typedef long long ll; const int IN ...
随机推荐
- AngularJS学习篇(五)
AngularJS Scope(作用域) Scope(作用域) 是应用在 HTML (视图) 和 JavaScript (控制器)之间的纽带. Scope 是一个对象,有可用的方法和属性. Scope ...
- 基于Django的python验证码
验证码 在用户注册.登录页面,为了防止暴力请求,可以加入验证码功能,如果验证码错误,则不需要继续处理,可以减轻一些服务器的压力 使用验证码也是一种有效的防止crsf的方法 验证码效果如下图: 验证码视 ...
- 你有什么理由还不选择阿里云服务器呢--从阿里云发布自研商用关系型数据库POLARDB想到的
最近几天,阿里云发布自研商用关系型数据库POLARDB的消息可谓是重磅炸弹啊.借用官方宣传的话就是:6倍性能于MySQL并100%兼容/100TB存储容量/2分钟创建只读副本/3分钟创建容灾实例,第三 ...
- 清理win10过期补丁的命令
作用是删除已经被新版本取代的旧系统文件 DISM.exe /Online /Cleanup-Image /StartComponentCleanup /ResetBase 注1: 执行后, 补丁就无法 ...
- 我是如何理解Android的Handler模型_3
AsyncTask则相当于现代化的电话系统,接线员的功能被完全封装了. 对于上例,新建更新TextView的类并继承AsyncTack类,如下: class UpdataTV extends Asyn ...
- linux操作系统基础篇(二)
Linux用户.群组.权限 1.用户也是由一个个文件组成的下列文件都是存放用户信息的文件 useradd user1 /etc/passwd: 存放用户信息 /etc/shadow/ :存放用户密码 ...
- github设置
ssh-key: https://help.github.com/articles/generating-ssh-keys http://segmentfault.com/q/101000000013 ...
- 三种读取HashMap的方式
package com.biubiu.entity; import java.util.Collection; import java.util.HashMap; import java.util.I ...
- Java并发之线程管理(线程基础知识)
因为书中涵盖的知识点比较全,所以就以书中的目录来学习和记录.当然,学习书中知识的时候自己的思考和实践是最重要的.说到线程,脑子里大概知道是个什么东西,但很多东西都还是懵懵懂懂,这是最可怕的.所以想着细 ...
- linux使用yum安装mariadb
一,安装 yum install mariadb mariadb-server 二,如何设置密码 用root 进入mysql后 mysql>set password =password('你的 ...