H - Lazier Salesgirl

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 3607

Description

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

2
4
1 2 3 4
1 3 6 10
4
4 3 2 1
1 3 6 10

Sample Output

4.000000 2.500000
1.000000 4.000000

题目大意就是一个买面包的小姑凉想偷懒，在卖到第i个人的时候，就会在一个之前维持的最大间隔时间内（如果在这个时间内没人来买面包的）她会一睡不醒。但又要满足能否去到最大平均值。
总体来说，这道题有两个条件：平均值最大，并且能一睡不醒。

#include<cstdio>
#include<string.h>
using namespace std;
double p[1010];
double time[1010];
double maxt[1010];
double max(double a,double b)
{
    return a>b?a:b;
}
int main()
{
    int t,n;
    double w;
    double flag;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%lf",&p[i]);
        for(int i=1;i<=n;i++)
            scanf("%lf",&time[i]);
        time[0]=0;
        maxt[1]=time[1]-time[0];
        for(int i=2;i<=n;i++)
            maxt[i]=max(time[i]-time[i-1],maxt[i-1]);//算出到第i个人时的前面的最大间隔时间
        double maxn=0;//最大平均值
        double anst=0;//间隔时间
        double sum=0;
        for(int i=1;i<=n;i++)//暴力枚举
        {
             w=maxt[i];
            sum+=p[i];
            if(i==n)
            {
                if(sum/i>maxn)
                {
                    maxn=sum/i;
                    flag=w;
                    break;
                }
            }
            if(sum/i>maxn&&w<time[i+1]-time[i])//要保证他卖给第i个人后能睡觉
            {
                maxn=sum/i;
                flag=w;
            }

        }
        printf("%.6lf %.6lf\n",flag,maxn);
    }
    return 0;
}