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.

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.

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.

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

Case #1: 3.333333
Case #2: 13.333333

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#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;
}