LightOJ-1079-Just another Robbery(概率, 背包)
链接:
https://vjudge.net/problem/LightOJ-1079#author=feng990608
题意:
As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants everything quick!) so he decided to rob banks. He wants to make a calculated risk, and grab as much money as possible. But his friends - Hermione and Ron have decided upon a tolerable probability P of getting caught. They feel that he is safe enough if the banks he robs together give a probability less than P.
思路:
概率,把小于最大值,变成大于最小概率,用背包搞一下就行,之前刷背包刷到过,现在居然忘了..
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
double Dp[100010], P[110];
int M[110];
int n;
int main()
{
int t, cnt = 0;
scanf("%d", &t);
while (t--)
{
memset(Dp, 0, sizeof(Dp));
double p;
int sum = 0;
scanf("%lf%d", &p, &n);
for (int i = 1;i <= n;i++)
scanf("%d%lf", &M[i], &P[i]), sum += M[i];
p = 1-p;
Dp[0] = 1.0;
for (int i = 1;i <= n;i++)
{
for (int j = sum;j >= M[i];j--)
Dp[j] = max(Dp[j], Dp[j-M[i]]*(1-P[i]));
}
int res = 0;
for (int i = sum;i >= 0;i--)
if (Dp[i] >= p)
{
res = i;
break;
}
printf("Case %d: %d\n", ++cnt, res);
}
return 0;
}
LightOJ-1079-Just another Robbery(概率, 背包)的更多相关文章
- LightOJ 1079 Just another Robbery 概率背包
Description As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (h ...
- LightOJ - 1079 Just another Robbery —— 概率、背包
题目链接:https://vjudge.net/problem/LightOJ-1079 1079 - Just another Robbery PDF (English) Statistics ...
- LightOJ 1079 Just another Robbery (01背包)
题意:给定一个人抢劫每个银行的被抓的概率和该银行的钱数,问你在他在不被抓的情况下,能抢劫的最多数量. 析:01背包,用钱数作背包容量,dp[j] = max(dp[j], dp[j-a[i] * (1 ...
- LightOJ 1079 Just another Robbery (01背包)
题目链接 题意:Harry Potter要去抢银行(wtf???),有n个银行,对于每个银行,抢的话,能抢到Mi单位的钱,并有pi的概率被抓到.在各个银行被抓到是独立事件.总的被抓到的概率不能超过P. ...
- lightoj 1079 Just another Robbery
题意:给出银行的个数和被抓概率上限.在给出每个银行的钱和抢劫这个银行被抓的概率.求不超过被抓概率上线能抢劫到最多的钱. dp题,转移方程 dp[i][j] = min(dp[i-1][j] , dp[ ...
- (概率 01背包) Just another Robbery -- LightOJ -- 1079
http://lightoj.com/volume_showproblem.php?problem=1079 Just another Robbery As Harry Potter series i ...
- 1079 - Just another Robbery
1079 - Just another Robbery PDF (English) Statistics Forum Time Limit: 4 second(s) Memory Limit: 3 ...
- LightOJ - 1079 概率dp
题意:n个银行,每个有价值和被抓概率,要求找被抓概率不超过p的最大价值 题解:dp[i][j]表示前i个取j价值的所需最小概率,01背包处理,转移方程dp[i][j]=min(dp[i-1][j],d ...
- hdu 2955 Robberies(概率背包)
Robberies Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total S ...
随机推荐
- KVM虚拟化储存管理(3)
一.KVM 存储虚拟化介绍 KVM 的存储虚拟化是通过存储池(Storage Pool)和卷(Volume)来管理的. Storage Pool 是宿主机上可以看到的一片存储空间,可以是多种型: Vo ...
- 白盒测试笔记之:testng 单元测试
前言 前一篇文章我们简单了解了下单元测试的概念以及使用junit进行入门了. 但想更好做自动化测试,还是得了解下testng,毕竟,作为一名技术人,NG(下一代)的测试框架总得了解与跟进. testn ...
- Leetcode之广度优先搜索(BFS)专题-752. 打开转盘锁(Open the Lock)
Leetcode之广度优先搜索(BFS)专题-752. 打开转盘锁(Open the Lock) BFS入门详解:Leetcode之广度优先搜索(BFS)专题-429. N叉树的层序遍历(N-ary ...
- __setattr__,__getattr__,__delattr__
class Foo: x = 1 def __init__(self,y): self.y = y def __getattr__(self,item): print("---->fr ...
- 解决incorrect 'only available in ES6' warning (W119) with options `moz: true, esversion: 6` 报错问题
很多同学在新建vue项目时,会遇到 incorrect 'only available in ES6' warning (W119) with options `moz: true, esversio ...
- 简单的利用nginx部署前端项目
网上有很多教程写的一大堆东西,新手可能会有点看不懂,现在我写这篇文章是为了更好的帮助新手,如何将自己的前端项目部署到自己的服务器上. 首先我们必须要有一台自己的ubuntu服务器,如果没有可以去阿里云 ...
- springboot_redis
1.引入redis的启动器 <dependency> <groupId>org.springframework.boot</groupId> <artifac ...
- python random 的用法
python random的里面的方法其实是Random实例化的对象. 里面几个常用的几个方import random print( random.randint(1,10) ) # 产生 1 到 1 ...
- Swool的安装与使用
1.swoole的安装 //php最好用7.2以上的.直接去网站下载下来,然后与php一样编译安装. git下来后,因为没有config文件,故先在swool下载目录下执行: /.../php/bin ...
- [HDU517] 小奇的集合
题目链接 显然有贪心每次选择最大的两个数来做. 于是暴力地把最大的两个数调整到非负(暴力次数不超过1e5),接下来使用矩阵乘法即可. \[ \begin{pmatrix} B'\\S'\\T' \en ...