Delivery

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

 

Source

2014 ACM/ICPC Asia Regional Beijing Online

 

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 题意：

#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=100010;
typedef long long ll;

int main()
{
    int t,cas=1,n,c,i;
    double k,sk;

    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        sk=0;
        for(i=1;i<=n;i++)
        {
            scanf("%lf",&k);
            sk+=k;
        }
        for(i=1;i<=n;i++)
            scanf("%d",&c);
        printf("Case #%d: %.6f\n",cas++,(n+1)/sk);
    }
    return 0;
}