hdu 5035 Delivery（北京网络赛）

最新推荐文章于 2019-08-07 18:54:00 发布

AC_way

最新推荐文章于 2019-08-07 18:54:00 发布

阅读量835

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM-区域赛文章标签： algorithm

本文链接：https://blog.youkuaiyun.com/caduca/article/details/39453711

ACM-区域赛专栏收录该内容

22 篇文章

订阅专栏

本文探讨了一个关于邮包送达的问题，具体来说，Matt需要将一个包裹送到Ted手中，在邮局遇到N位正在为当前顾客服务的职员，每位职员的服务时间服从特定的指数分布，通过给定的职员效率和已服务时间，计算Matt完成邮寄所需的平均等待时间。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Delivery

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 41 Accepted Submission(s): 21

Problem Description

Today, Matt goes to delivery a package to Ted. When he arrives at the post office, he finds there are N clerks numbered from 1 to N. All the clerks are busy but there is no one else waiting in line. Matt will go to the first clerk who has finished the service for the current customer. The service time t_i for clerk i is a random variable satisfying distribution p(t_i = t) = k_ie^(-k_it) where e represents the base of natural logarithm and k_i represents the efficiency of clerk i. Besides, accroding to the bulletin board, Matt can know the time c_i which clerk i has already spent on the current customer. Matt wants to know the expected time he needs to wait until finishing his posting given current circumstances.

Input

The first line of the input contains an integer T,denoting the number of testcases. Then T test cases follow.

For each test case, the first line contains one integer:N(1<=N<=1000)

The second line contains N real numbers. The i-th real number k_i(0<=k_i<=1) indicates the efficiency of clerk i.

The third line contains N integers. The i-th integer indicates the time c_i(0<=c_i<=1000) which clerk i has already spent on the current customer.

Output

For each test case, output one line "Case #x: y", where x is the case number (starting from 1), y is the answer which should be rounded to 6 decimal places.

Sample Input

2
2
0.5 0.4
2 3
3
0.1 0.1 0.1
3 2 5

Sample Output

Case #1: 3.333333
Case #2: 13.333333

推出公式就好了，c没用。

代码：

//453ms
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int main()
{
    int t;
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
        int n;
        double ks=0;
        double x;
        int c;
        scanf("%d",&n);
        for(int j=1;j<=n;j++)
        {
            scanf("%lf",&x);
            ks+=x;
        }
        for(int j=1;j<=n;j++)
        scanf("%d",&c);
        printf("Case #%d: ",i);
        printf("%.6f\n",(n+1)*1.0/ks);
    }
    return 0;
}