ZOJ3607-Lazier Salesgirl

Kochiya Sanae is a lazy girl who makes and sells bread. She is an expert at bread making and selling. She can sell the i-th customer a piece of bread for PRice p_i. But she is so lazy that she will fall asleep if no customer comes to buy bread for more than w minutes. When she is sleeping, the customer coming to buy bread will leave immediately. It's known that she starts to sell bread now and the i-th customer come after t_i minutes. What is the minimum possible value of w that maximizes the average value of the bread sold?

Input

There are multiple test cases. The first line of input is an integer T ≈ 200 indicating the number of test cases.

The first line of each test case contains an integer 1 ≤ n ≤ 1000 indicating the number of customers. The second line contains n integers 1 ≤ p_i ≤ 10000. The third line contains n integers 1 ≤ t_i ≤ 100000. The customers are given in the non-decreasing order of t_i.

Output

For each test cases, output w and the corresponding average value of sold bread, with six decimal digits.

Sample Input

241 2 3 41 3 6 1044 3 2 11 3 6 10
Sample Output
4.000000 2.5000001.000000 4.000000
Author: WU, ZejunContest: The 9th Zhejiang Provincial Collegiate Programming Contest
题意：一个卖面包的小姑娘，给第i个来买面包的人的价格是pi，每个人的时间是ti，过了时间t还没人来买她就会一睡不起，不然她一直会卖面包，求最小的时间间隔，最大的平均值。
#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>using namespace std;const int INF=0x3f3f3f3f;int p[10009],t[10009];int main(){    int T,n;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)            scanf("%d",&p[i]);        for(int i=1;i<=n;i++)            scanf("%d",&t[i]);        double mit,ma=-INF,sum=0,zt=-INF;        t[n+1]=INF;        for(int i=1;i<=n;i++)        {            zt=max(1.0*t[i]-t[i-1],zt);            sum+=p[i];            double k=sum/i;            if(k>ma&&t[i+1]-t[i]>zt)            {                ma=k;                mit=zt;            }        }        printf("%.6lf %.6lf/n",mit,ma);    }    return 0;}