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 nintegers 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, Zejun
Contest: The 9th Zhejiang Provincial Collegiate Programming Contest

题意：

T个情况，每个情况有n个客人，第i个客人可赚p[i]元钱，第i个客人t[i]时间来，卖东西的女孩很懒，如果w时间内没人来就睡觉，后面来的客人就会刚来就走，求赚钱的平均值最大同时输出最小w.

这题采用贪心，如果t[i+1]-t[i]<maxtime时，那么后面一个客人能买到面包，如果t[i+1]-t[i]>maxtime那么就先把当前的情况记录下来，再往后去找

 1 #include<cstdio> 2 #include<cstring> 3 #include<stdlib.h> 4 #include<algorithm> 5 using namespace std; 6 const int MAXN=1000+10; 7 int p[MAXN],t[MAXN]; 8 int main() 9 {10     //freopen("in.txt","r",stdin);11     int kase;12     scanf("%d",&kase);13     while(kase--)14     {15         int n;16         scanf("%d",&n);17         memset(p,0,sizeof(p));18         memset(t,0,sizeof(t));19         for(int i=1;i<=n;i++)20             scanf("%d",&p[i]);21         for(int i=1;i<=n;i++)22             scanf("%d",&t[i]);23 24         double maxtime=-1,time=0;25         double sum=0,av=0;26         for(int i=1;i<=n;i++)27         {28             sum+=p[i];29             if(maxtime<t[i]-t[i-1])30                 maxtime=t[i]-t[i-1];31 32             if(maxtime<t[i+1]-t[i]&&sum>av*i)33             {34                 av=sum/i;35                 time=maxtime;36             }37 38             if(i==n)39             {40                 if(av*i<sum)41                 {42                     av=sum/i;43                     time=maxtime;44                 }45             }46         }47         printf("%.6lf %.6lf\n",time,av);48     }49     return 0;50 }

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