HDU 2955 【01背包+小数概率】
Robberies
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29495 Accepted Submission(s): 10795
Problem Description
The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usually gets caught in the end, often because they become too greedy. He has decided to work in the lucrative business of bank robbery only for a short while, before retiring to a comfortable job at a university.
For a few months now, Roy has been assessing the security of various banks and the amount of cash they hold. He wants to make a calculated risk, and grab as much money as possible.
His mother, Ola, has decided upon a tolerable probability of getting caught. She feels that he is safe enough if the banks he robs together give a probability less than this.
Input
The first line of input gives T, the number of cases. For each scenario, the first line of input gives a floating point number P, the probability Roy needs to be below, and an integer N, the number of banks he has plans for. Then follow N lines, where line j gives an integer Mj and a floating point number Pj .
Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj .
Output
For each test case, output a line with the maximum number of millions he can expect to get while the probability of getting caught is less than the limit set.
Notes and Constraints
0 < T <= 100
0.0 <= P <= 1.0
0 < N <= 100
0 < Mj <= 100
0.0 <= Pj <= 1.0
A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.
Sample Input
3
0.04 3
1 0.02
2 0.03
3 0.05
0.06 3
2 0.03
2 0.03
3 0.05
0.10 3
1 0.03
2 0.02
3 0.05
Sample Output
2
4
6
Source
IDI Open 2009
HDU 2955 【01背包+小数概率】的更多相关文章
- HDU 2955 【01背包/小数/概率DP】
Robberies Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Sub ...
- hdu 2955 01背包
http://acm.hdu.edu.cn/showproblem.php?pid=2955 如果认为:1-P是背包的容量,n是物品的个数,sum是所有物品的总价值,条件就是装入背包的物品的体积和不能 ...
- HDU 1203 【01背包/小数/概率DP】
I NEED A OFFER! Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Tot ...
- HDU 2955 01背包(思维)
Robberies Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total S ...
- Robberies hdu 2955 01背包
Robberies Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total S ...
- hdu 1203 01背包 I need a offer
hdu 1203 01背包 I need a offer 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1203 题目大意:给你每个学校得到offe ...
- HDU 2955 Robberies 背包概率DP
A - Robberies Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submi ...
- HDU 1203 I NEED A OFFER!(01背包+简单概率知识)
I NEED A OFFER! Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Sub ...
- HDU 1203 I NEED A OFFER! (动态规划、01背包、概率)
I NEED A OFFER! Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)T ...
随机推荐
- 实际遭遇GC回收造成的Web服务器CPU跑高
今天下午有段时间访问园子感觉不如以前那么快的流畅,上Web服务器一看,果然,负载均衡中的1台云服务器CPU跑高. 上图中红色曲线表示的是CPU占用率.正常情况下,CPU占用率一般在40%以下. 这台云 ...
- WebDriver--简单元素操作
clear():清除文本,可用来键盘输入前清除一些input输入框默认的值 send_key(*value):模拟按键输入 click():单击,不止按钮,也可以是文字/图片链接.复选框.单选框.下拉 ...
- IDEA调试快捷键
F9 resume programe 恢复程序 F8 Step Over 相当于eclipse的f6 跳到下一步 Ctrl+Shift+F,全局查 ...
- mysql安装目录、配置文件存放位置
linux系统下,如何知道mysql使用的配置文件到底是哪个呢?linux自带的mysql的安装目录又是什么呢?数据存放在什么目录下? 1.linux系统自带的mysql,其安装目录及数据目录查看方法 ...
- [常识]Windows系统里休眠和睡眠的区别?
睡眠和休眠都是笔记本电脑的节能方式,但有细微的差别: 睡眠还保持着开机状态的,休眠是关机了,但是再次开机之后和关闭时的系统状态是一样的. 睡眠还是保持着系统运行数据在内存中,而休眠则将内存中的数据保存 ...
- select chosen 禁用下拉框某一个option
$("#tbParBudCode option[value='" + budCodeId + "']").attr("disabled", ...
- P4305 [JLOI2011]不重复数字
题目描述 给出N个数,要求把其中重复的去掉,只保留第一次出现的数. 例如,给出的数为1 2 18 3 3 19 2 3 6 5 4,其中2和3有重复,去除后的结果为1 2 18 3 19 6 5 4. ...
- 架构-UML类图
在UML 2.0的13种图形中,类图是使用频率最高的UML图之一.Martin Fowler在其著作<UML Distilled: A Brief Guide to the Standard O ...
- 深入解析vue.js响应式原理与实现
vue.js响应式原理解析与实现.angularjs是通过脏检查来实现数据监测以及页面更新渲染.之后,再接触了vue.js,当时也一度很好奇vue.js是如何监测数据更新并且重新渲染页面.vue.js ...
- [poj] 1375 Interval || 圆的切线&和直线的交点
原题 每组数据给出一些圆(障碍物)的圆心和半径,一个点和一条线段,求站在这个点,能开到的线段的部分的左端点和右端点.没有则输出"No View" 相当于求过该点的圆的两条切线,切线 ...