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。

 1 #include<cstdio>
 2 #include<string.h>
 3 using namespace std;
 4 double p[1010];
 5 double time[1010];
 6 double maxt[1010];
 7 double max(double a,double b)
 8 {
 9     return a>b?a:b;
10 }
11 int main()
12 {
13     int t,n;
14     double w;
15     double flag;
16     scanf("%d",&t);
17     while(t--)
18     {
19         scanf("%d",&n);
20         for(int i=1;i<=n;i++)
21             scanf("%lf",&p[i]);
22         for(int i=1;i<=n;i++)
23             scanf("%lf",&time[i]);
24         time[0]=0;
25         maxt[1]=time[1]-time[0];
26         for(int i=2;i<=n;i++)
27             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
28         double maxn=0;//最大平均值
29         double anst=0;//间隔时间
30         double sum=0;
31         for(int i=1;i<=n;i++)//暴力枚举
32         {
33              w=maxt[i];
34             sum+=p[i];
35             if(i==n)
36             {
37                 if(sum/i>maxn)
38                 {
39                     maxn=sum/i;
40                     flag=w;
41                     break;
42                 }
43             }
44             if(sum/i>maxn&&w<time[i+1]-time[i])//要保证他卖给第i个人后能睡觉
45             {
46                 maxn=sum/i;
47                 flag=w;
48             }
49
50         }
51         printf("%.6lf %.6lf\n",flag,maxn);
52     }
53     return 0;
54 }

H - Lazier Salesgirl