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[];
 double time[];
 double maxt[];
 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=;i<=n;i++)
             scanf("%lf",&p[i]);
         for(int i=;i<=n;i++)
             scanf("%lf",&time[i]);
         time[]=;
         maxt[]=time[]-time[];
         for(int i=;i<=n;i++)
             maxt[i]=max(time[i]-time[i-],maxt[i-]);//算出到第i个人时的前面的最大间隔时间
         double maxn=;//最大平均值
         double anst=;//间隔时间
         double sum=;
         for(int i=;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+]-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 ;
 }