zoukankan      html  css  js  c++  java
  • Hdu 2955 Robberies

    Robberies

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 23934    Accepted Submission(s): 8842

    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

    #include <stdio.h>  
    #include <algorithm>  
    #include <iostream>  
    #include <cmath>  
    #include <string.h>  
    using namespace std;  
      
    const int N = 50005;  
      
    int main()  
    {  
      //  freopen("in.txt", "r", stdin);      
    int t, m0, m[N];  
        double p0, p[N], ans[N];  
        scanf("%d", &t);  
        while(t --)  
        {  
            scanf("%lf%d", &p0, &m0);  
            int sum = 0;  
            for(int i = 0; i < m0; i ++)  
            {  
                scanf("%d%lf", &m[i], &p[i]);  
                sum += m[i];  
            }  
            memset(ans, 0, sizeof(ans));  
            ans[0] = 1;  
            for(int i = 0; i < m0; i ++)  
            {  
                for(int j = sum; j >= m[i]; j --)  
                {  
                    ans[j] = max(ans[j], ans[j - m[i]] *(1 - p[i]));  
                }  
            }  
            for(int i = sum; i >= 0; i --)  
            {  
                if(ans[i] > (1 - p0))  
                {  
                    printf("%d
    ", i);  
                    break;  
                }  
            }  
        }  
        return 0;  
    }  
    

      

  • 相关阅读:
    [HNOI2016]序列
    [Cqoi2015] 编号 【逆向思维,暴力枚举】
    [NOI2015] 软件包管理器【树链剖分+线段树区间覆盖】
    [Hdu-6053] TrickGCD[容斥,前缀和]
    [Hdu-5155] Harry And Magic Box[思维题+容斥,计数Dp]
    牛客NOIP暑期七天营-提高组6
    [SHOI2007] 书柜的尺寸 思维题+Dp+空间优化
    [UVA12235] Help Bubu 思维题+状态定义+Dp
    牛客NOIP暑期七天营-TG3 赛后题解
    牛客NOIP暑期七天营-TG1 赛后题解
  • 原文地址:https://www.cnblogs.com/zhangliu/p/7063221.html
Copyright © 2011-2022 走看看