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
- 将有效慢日志转存到数据库v2
import re import sys import getopt import MySQLdb from subprocess import call import os host='10.76. ...
- 【Validation】林轩田机器学习基石
这一节主要讲如何通过数据来合理的验证模型好不好. 首先,否定了Ein来选模型和Etest来选模型. (1)模型越复杂,Ein肯定越好:但是Eout就不一定了(见上一节的overfitting等) (2 ...
- python 学习分享-paramiko模块
paramiko模块学习分享 paramiko是用python语言写的一个模块,遵循SSH2协议,支持以加密和认证的方式,进行远程服务器的连接.paramiko支持Linux, Solaris, BS ...
- Singleton patterns 单件(创建型模式)
1.模式分类 1.1 从目的来看: • – 创建型(Creational)模式:负责对象创建. • – 结构型(Structural)模式:处理类与对象间的组合. • ...
- 孤荷凌寒自学python第四十四天Python操作 数据库之准备工作
孤荷凌寒自学python第四十四天Python操作数据库之准备工作 (完整学习过程屏幕记录视频地址在文末,手写笔记在文末) 今天非常激动地开始接触Python的数据库操作的学习了,数据库是系统化设计 ...
- CSU-1986 玄学
题目链接 http://acm.csu.edu.cn:20080/csuoj/problemset/problem?pid=1986 题目 Description 阴阳师子浩君,最近从<初等数论 ...
- 【志银】Dev-Cpp配置OpenGL图形库(成功版本:Dev-Cpp 5.7.1 MinGW 4.8.1)
★配置前须知:Dev-Cpp自带OpenGL的使用和OpenGL简介 (附Dev-Cpp下载地址:http://sourceforge.net/projects/orwelldevcpp/?sourc ...
- HDU 4665 Unshuffle DFS找一个可行解
每层找一对相等的整数,分别放在两个不同的串中. 参考了学弟的解法,果断觉得自己老了…… #include <cstdio> #include <cstring> #includ ...
- perror表
#define EPERM 1 /* Operation not permitted */ #define ENOENT 2 /* No such file or directory */ #defi ...
- C++ 虚函数的内存分配
1.无继承的普通类: 在有虚函数的情况下类会为其增加一个隐藏的成员,虚函数表指针,指向一个虚函数表,虚函数表里面就是类的各个虚函数的地址了.那么,虚函数表指针是以什么模型加入到类里面的,虚函数表里 ...