1079 - Just another Robbery
Time Limit: 4 second(s) Memory Limit: 32 MB

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.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains a real number P, the probability Harry needs to be below, and an integer N (0 < N ≤ 100), the number of banks he has plans for. Then follow N lines, where line j gives an integer Mj (0 < Mj ≤ 100) and a real number Pj . Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj. A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.

Output

For each case, print the case number and the maximum number of millions he can expect to get while the probability of getting caught is less than P.

Sample Input

Output for 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

Case 1: 2

Case 2: 4

Case 3: 6

Note

For the first case, if he wants to rob bank 1 and 2, then the probability of getting caught is 0.02 + (1 - 0.02) * .03 = 0.0494 which is greater than the given probability (0.04). That's why he has only option, just to rob rank 2.


Problem Setter: Jane Alam Jan
思路:01背包;
被抓的概率不容易求,那么转变为不被抓的概率,去求,然后因为钱的范围不大,所以按钱来做一个01背包。
 1 #include<stdio.h>
2 #include<algorithm>
3 #include<iostream>
4 #include<stdlib.h>
5 #include<queue>
6 #include<string.h>
7 using namespace std;
8 typedef struct node
9 {
10 int cost;
11 double pr;
12 } ss;
13 ss ans[105];
14 double dp[10005];
15 int main(void)
16 {
17 int T;
18 scanf("%d",&T);
19 int __ca = 0;
20 double pi;
21 while(T--)
22 {
23 __ca++;
24 int n;int sum = 0;
25 scanf("%lf %d",&pi,&n);
26 int i,j;pi = 1-pi;
27 memset(dp,0,sizeof(dp));
28 for(i = 1; i <= n; i++)
29 {
30 scanf("%d %lf",&ans[i].cost,&ans[i].pr);
31 ans[i].pr = 1-ans[i].pr;
32 sum += ans[i].cost;
33 }
34 dp[0] = 1;
35 for(i = 1;i <= n;i++)
36 {
37 for(j = sum;j >= ans[i].cost; j--)
38 {
39 dp[j] = max(dp[j],dp[j-ans[i].cost]*ans[i].pr);
40 }
41 }
42 int ask = 0;
43 for(i = 1;i <= sum ;i++)
44 {
45 if(dp[i] >= pi)
46 {
47 ask = i;
48 }
49 }
50 printf("Case %d: %d\n",__ca,ask);
51 }
52 return 0;
53 }

1079 - Just another Robbery的更多相关文章

  1. 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 ...

  2. 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 ...

  3. 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[ ...

  4. 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 ...

  5. 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. ...

  6. (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 ...

  7. Just another Robbery(背包)

    1079 - Just another Robbery   PDF (English) Statistics Forum Time Limit: 4 second(s) Memory Limit: 3 ...

  8. Light OJ Dynamic Programming

    免费做一样新 1004 - Monkey Banana Problem 号码塔 1005 - Rooks 排列 1013 - Love Calculator LCS变形 dp[i][j][k]对于第一 ...

  9. KUANGBIN带你飞

    KUANGBIN带你飞 全专题整理 https://www.cnblogs.com/slzk/articles/7402292.html 专题一 简单搜索 POJ 1321 棋盘问题    //201 ...

随机推荐

  1. Mybatis相关知识点(一)

    MyBatis入门 (一)介绍 MyBatis 本是apache的一个开源项目iBatis, 2010年这个项目由apache software foundation 迁移到了google code, ...

  2. 商业爬虫学习笔记day5

    一. 发送post请求 import requests url = "" # 发送post请求 data = { } response = requests.post(url, d ...

  3. vim中搜索指定单词(不加前后缀)

    \< : 搜索内容作为单词开头 \> : 搜索内容作为单词结尾 一起用即为将搜索内容指定为whole word e.g. : word_suffix word 如果用/word来搜索则两个 ...

  4. springmvc中拦截器的定义和配置

    package com.hope.interceptor;import org.springframework.lang.Nullable;import org.springframework.web ...

  5. bugku 杂项 流量分析(cnss)

    bugku 杂项 流量分析(cnss) 此题较为简单 wireshark 追踪第一行tcp流信息 得到如下 GET /stat.htm?id=2724999&r=http%3A%2F%2Fsp ...

  6. 【Xcode】sh: pause: command not found

    system("pause"); 只适合于DOS和Windows系统,不适合Linux系统. 直接删掉就可以. 或者改为: #include <unistd.h> pa ...

  7. SQLserver 2014使用Convert()函数获取时间

    select convert(char(100),GetDate(),120) as Date 第3个参数就是用来设置日期类型数据的显示样式的,下面介绍几种样式的参数 SELECT CONVERT(v ...

  8. vue文件上传及压缩(canvas实现压缩)

    // 读取文件结果 afterRead(files) { let that = this; let file = files.file; if (file === undefined) { retur ...

  9. NTLM验证过程

    中我们介绍Kerberos认证的整个流程.在允许的环境下,Kerberos是首选的认证方式.在这之前,Windows主要采用另一种认证协议 --NTLM(NT Lan Manager).NTLM使用在 ...

  10. [Java Web 王者归来]读书笔记1

    第一章 Java web 开发概述 1 WEB服务器运行时一直在TCP 80(默认端口)监听, 若使用其他端口在url中需要显示标注端口号(例如:8080) 2 WEB服务器:微软的IIS.Apach ...