acdream1421 TV Show (枚举)
http://acdream.info/problem?pid=1421
Andrew Stankevich Contest 22
| TV ShowSpecial JudgeTime Limit: 2000/1000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others) Problem Description      Charlie is going to take part in one famous TV Show. The show is a single player game. Initially the player has 100 dollars. The player is asked n questions, one after another. If the player answers the question correctly, the sum he has is doubled. If the answer is incorrect, the player gets nothing and leaves the show.       Before each question the player can choose to leave the show and take away the prize he already has. Also once in the game the player can buy insurance. Insurance costs c dollars which are subtracted from the sum the player has before the question. Insurance has the following effect: if the player answers the question correctly his prize is doubled as usually, if the player answers incorrectly the prize is not doubled, but the game continues. The player must have more than c dollars to buy insurance.       Charlie’s friend Jerry works on TV so he managed to steal the topics of the questions Charlie will be asked. Therefore for each question i Charlie knows pi — the probability that he will answer this question correctly. Now Charlie would like to develop the optimal strategy to maximize his expected prize. Help him. Input      The first line of the input file contains two integer numbers n and c (1 ≤ n ≤ 50, 1 ≤ c ≤ 109). The second line contains n integer numbers ranging from 0 to 100 — the probabilities that Charlie will answer questions correctly, in percent. Output      Output one real number — the expected prize of Charlie if he follows the optimal strategy. Your answer must have relative or absolute error within 10-8 . Sample Input2 100 Sample Output100 Hint      The optimal strategy in the second example is to take insurance for the second question. In this case the expected prize is 1/2 × 0 + 1/2 × (1/2 × 150 + 1/2 × 300).       In the third example it is better to leave the show after the first question, because there is no reason to try to answer the second one. SourceAndrew Stankevich Contest 22 Manager | 
题意:
有个人参加答题活动,初始有100块,答对一题翻倍,答错一题直接没钱滚蛋。他可以选择在任意时刻拿着现有的钱跑。已知各个题答对的概率,求最高的得钱期望。
还有一个特殊设定,就是他可以在某个时刻买一个保险,保险需要c元,买了的话下一题答错了也不会滚蛋,而是会保持原有钱数继续。
题解:
注意保险只能买一次,直接枚举就行!
(之前以为保险随便买,卧槽,就不会了)
代码:
//#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<cmath>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
#include<stack>
#include<queue>
using namespace std;
#define ll long long
#define usll unsigned ll
#define mz(array) memset(array, 0, sizeof(array))
#define mf1(array) memset(array, -1, sizeof(array))
#define minf(array) memset(array, 0x3f, sizeof(array))
#define REP(i,n) for(i=0;i<(n);i++)
#define FOR(i,x,n) for(i=(x);i<=(n);i++)
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define WN(x) printf("%d\n",x);
#define RE freopen("D.in","r",stdin)
#define WE freopen("huzhi.txt","w",stdout)
#define mp make_pair
#define pb push_back
const double pi=acos(-1.0);
const double eps=1e-; const int maxn=; int n;
double c;
int a[maxn]; inline double dfs(const int &x,const double &s,const bool bought) {
double t=s;
if(x>=n)return s;
t=max(t , a[x]/100.0*dfs(x+ , *s , bought));
if(s>=c && bought!=)t=max(t , dfs(x+ , (a[x]*2.0*(s-c) + (-a[x])*(s-c))/100.0 , ));
return t;
} double farm() {
return dfs(,100.0,);
} int main() {
int i;
while(scanf("%d%lf",&n,&c)!=EOF) {
REP(i,n)RD(a[i]);
printf("%f\n",farm());
}
}
acdream1421 TV Show (枚举)的更多相关文章
- UVA1660 电视网络 Cable TV Network
		题目地址:UVA1660 电视网络 Cable TV Network 枚举两个不直接连通的点 \(S\) 和 \(T\) ,求在剩余的 \(n-2\) 个节点中最少去掉多少个可以使 \(S\) 和 \ ... 
- uva140-暴力枚举
		题意:任意一个点都至少有一个点与其相连接,所有的点可以进行任意排列,总排列数为n!. 一个点带宽定义与它相连的点的最远距离,一个排列的带宽定义为,点中最大的带宽,找出带宽最小的那个排列,有多组,输出字 ... 
- C# - 集合类
		C#的集合类命名空间介绍: // 程序集 mscorlib.dll System.dll System.Core.dll // 命名空间 using System.Collections:集合的接口和 ... 
- POJ 1966 Cable TV Network
		Cable TV Network Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 4702 Accepted: 2173 ... 
- 使用EF自带的EntityState枚举和自定义枚举实现单个和多个实体的增删改查
		本文目录 使用EntityState枚举实现单个实体的增/删/改 增加:DbSet.Add = > EntityState.Added 标记实体为未改变:EntityState.Unchange ... 
- Cable TV Network-POJ1966图的连通度
		Cable TV Network Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 4404 Accepted: 2047 Desc ... 
- CSU OJ PID=1514: Packs   超大背包问题,折半枚举+二分查找。
		1514: Packs Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 61 Solved: 4[Submit][Status][Web Board] ... 
- C# 中的枚举类型 enum (属于值类型)
		原文 C# 中的枚举类型 enum (属于值类型) C# 支持两种特殊的值类型:枚举和结构. 声明枚举:声明时要声明所有可能的值. using System; using System.Collect ... 
- NYOJ 1091 超大01背包(折半枚举)
		这道题乍一看是普通的01背包,最最基础的,但是仔细一看数据,发现普通的根本没法做,仔细观察数组发现n比较小,利用这个特点将它划分为前半部分和后半部分这样就好了,当时在网上找题解,找不到,后来在挑战程序 ... 
随机推荐
- Hdfs常用操作
			一.linux rm是删除,不是del 二.常用操作 package hdfs; import java.io.FileInputStream; import java.io.FileNotFound ... 
- Linux解压命令(tar)
			tar zxvf phddns_raspberry.tgz tar默认解压到当前目录 参数: 运行tar时必须要有下列参数中的至少一个才可运行 -A, --catenate, --concatenat ... 
- Bzoj3524 [Poi2014]Couriers
			Description 给一个长度为n的序列a.1≤a[i]≤n. m组询问,每次询问一个区间[l,r],是否存在一个数在[l,r]中出现的次数大于(r-l+1)/2.如果存在,输出这个数,否则输出0 ... 
- bzoj4318OSU &tyvj1952 Easy
			之前做tyvj1952Easy(给定一个序列,每个位置有一定的概率是1或0,求极长连续1的长度平方期望),用的做法是求出"全1子串的期望个数".假如每一段极长连续1分别长x1,x2 ... 
- nagios note
			server: nagios nagios_plugin nrpe ip:192.168.1.2 client nagios_plugin xinetd nrpe ip:192.168.1.3 一篇介 ... 
- Android登录界面实现
			花了一些时间实现了一个还算可以等登陆界面,主要是对这两天工作的一个总结:自定义按钮.编辑框.布局.全屏等. 效果如下: 获取代码:点这里 
- 批处理(.bat)For命令使用
			for命令使用 @echo off REM 打印当前目录里所有子目录的名称 REM for /d %%i in (*) do @echo %%i REM 打印当前目录里所有名字为3个字符的子目录的名称 ... 
- 64位CentOS源码编译方式安装wine
			说明:本文仅作本人笔记的之用,仅供参考.可能因不同环境而不同. 1. 从官网下载最新版的wine-1.6.2.tar.gz 2. 安装相关的包(这里是我安装的,可能由于不同系统已经安装的包不同而不一样 ... 
- MOOCULUS微积分-2: 数列与级数学习笔记 1. Sequences
			此课程(MOOCULUS-2 "Sequences and Series")由Ohio State University于2014年在Coursera平台讲授. PDF格式教材下载 ... 
- UOJ#246. 【UER #7】套路
			题目传送门 官方题解传送门 一句话题意的话就是给定一个序列,从中找出至少$k$个连续的元素形成子序列,使得子序列中任意两个元素差值的最小值于其长度-1的乘积最大. 题目中给出了$ 1 \leq a_i ... 
