博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
Codeforces 442B
阅读量:7023 次
发布时间:2019-06-28

本文共 2871 字,大约阅读时间需要 9 分钟。

B. Andrey and Problem
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.

Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi(0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.

Output

Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.

Sample test(s)
Input
4 0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2 0.1 0.2
Output
0.260000000000
Note

In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.

In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.

 

Accepted Code:

1 /************************************************************************* 2     > File Name: 442B.cpp 3     > Author: Stomach_ache 4     > Mail: sudaweitong@gmail.com 5     > Created Time: 2014年06月23日 星期一 14时19分19秒 6     > Propose:  7  ************************************************************************/ 8  9 #include 
10 #include
11 #include
12 #include
13 #include
14 #include
15 #include
16 using namespace std;17 18 #define eps 1e-919 int n;20 double p[105];21 22 double cal(int x) {23 double result = 0.0;24 for (int i = x; i < n; i++) {25 double tmp = p[i];26 for (int j = x; j < n; j++) {27 if (i == j) continue;28 tmp *= (1.0 - p[j]);29 }30 result += tmp;31 }32 return result;33 }34 35 int36 main(void) {37 while (~scanf("%d", &n) && n) {38 for (int i = 0; i < n; i++) scanf("%lf", p+i);39 sort(p, p + n);40 double ans = p[n-1];41 for (int i = n-2; i >= 0; i--) {42 double tmp = cal(i);43 if (tmp - ans < eps) break;44 ans = tmp;45 }46 printf("%.17f\n", ans);47 }48 49 return 0;50 }

 

转载于:https://www.cnblogs.com/Stomach-ache/p/3805869.html

你可能感兴趣的文章
Java IO 之 FileInputStream & FileOutputStream源码分析
查看>>
时序列数据库武斗大会之什么是 TSDB ?
查看>>
fork() and vfork() getppid's result
查看>>
线程的其他特征
查看>>
《PostgreSQL 权威指南》Request For Comments - 结构
查看>>
【JSP开发】一个防盗链的WEB小例子
查看>>
我遇见的哪些 CSS 中有趣的尺寸、宽高
查看>>
域名分配问题
查看>>
VUE缓存:动态keep-alive
查看>>
杂项记录薄
查看>>
python 元类
查看>>
怎么让代码不再臃肿,写的像诗一样优雅
查看>>
移动端如果没有meta标签的viewport
查看>>
前端面试题
查看>>
WebSocket原理及技术简介
查看>>
自定义控件 --- 电池icon
查看>>
嘻哈说:设计模式之工厂方法模式
查看>>
JS原生Ajax基本操作
查看>>
JS == 操作符的隐式转换,翻译ecma-262/5.1/#sec-11.6.1
查看>>
大三学生的第二个基于 React 框架的轮播图组件。
查看>>