ZOJ 3607 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

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

贪心用map存储时间间隔。

只有两种情况：

如果后面的时间间隔比前面出现的最大的时间间隔还大的话，新建键值对存到map。

反之，在原来的最大时间间隔上进行处理。

 #include <stdio.h>
 #include <map>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 int n;
 int cnt;
 int sum[];
 map< double , double > M;
 map< double , double >::iterator it;

 struct Node{
     int p,t;
 }nod[];

 bool cmp(Node n1, Node n2){
     return n1.t<n2.t;
 }

 int main(){
     int t;
     int i,j;
     scanf("%d",&t);
     while( t-- ){
         scanf("%d" ,&n);
         for(i=; i<n; i++){
             scanf("%d" ,&nod[i].p);
         }
         for(i=; i<n; i++){
             scanf("%d" ,&nod[i].t);
         }
         sort(nod ,nod+n ,cmp);
         M.clear();
         for(i=; i<n; i++){
             if(i==){
                 sum[i]=nod[i].p;
             }else{
                 sum[i]=sum[i-]+nod[i].p;
             }
         }
         cnt=nod[].t;
         M[cnt]=nod[].p;
         for(i=; i<n; i++){
             if( nod[i].t-nod[i-].t>cnt ){
                 cnt=nod[i].t-nod[i-].t;
                 M[cnt]=double(sum[i])/(i+);
             }else{
                 M[cnt]=double(sum[i])/(i+);
             }
         }
         double w=;
         double maxValue=;
         for(it=M.begin(); it!=M.end(); it++){
             if( it->second > maxValue ){
                 w=it->first;
                 maxValue=it->second;
             }
         }
         printf("%.6lf %.6lf\n",w,maxValue);
     }
     return ;
 }