Lazier Salesgirl
Time Limit: 2 Seconds Memory Limit: 65536 KB
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 pi. 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 ti 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 ≤ pi ≤ 10000. The third line containsn integers 1 ≤ ti ≤
100000. The customers are given in the non-decreasing order of ti.
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
链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3607
题意：
某 在0时刻开店， 然后按顺序来了n个客人，分别出了一个价格p[i]，每个人来的时刻为 t[i]。
然后 某 在w时间内没接待客人就会睡觉，如果间隔正好是w，不会睡。 问接待的客人 出价的平均值最大是多少，在满足该条件下 w最低是多少。
做法：
不断求前缀和 还有 到第i个人间隔最大值。 接待的人肯定是连续的 前多少个。 先假设只接待第一个人，然后假设接待前两个，再前三个。 类推。
如果是正好前i个人，那么w得 大于等于 （ 到第i个人间隔最大值） ，而且要严格小于（第 i和i+1 个人的间隔差）。 所以w就 取 （ 到第i个人间隔最大值），
如果这个值严格小于 （第 i和i+1 个人的间隔差）那么就成立，判断前i个人的平均价格是否更低。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>
#define INF 999999999
#define eps 0.00001
#define LL __int64d
#define pi acos(-1.0)
struct node
{
int p, t;
int sum;
int big;//qian cha zui xiao
int cha;
}a[1047];
int cmp(node a,node b)
{
return a.sum>b.sum;
}
int main()
{
int t;
scanf("%d",&t);//xiao w da pingjun
while(t--)
{
int n, i, j;
double sum = 0;
scanf("%d",&n);
a[0].sum=0;
a[0].t=0;
for(i = 1; i <= n; i++)
{
scanf("%d",&a[i].p);
a[i].sum=a[i-1].sum+a[i].p;
}
for(i = 1; i <= n; i++)
{
scanf("%d",&a[i].t);
if(i==1)
a[i].big=a[i].t-a[i-1].t;
else
a[i].big=max(a[i-1].big,a[i].t-a[i-1].t);
}
for(i=1;i<=n;i++)
{
if(i==n)
a[i].cha=99999999;
else
a[i].cha=a[i+1].t-a[i].t;
}
//sort(a+1,a+1+n,cmp);
double bigavg=-1;
double big;
for(i=1;i<=n;i++)
{
if((1.0*a[i].sum/i)>bigavg)
{
if(a[i].cha>a[i].big)
{
big=a[i].big;
bigavg=(1.0*a[i].sum/i);
}
}
}
printf("%.6lf %.6lf\n",big,bigavg);
}
return 0;
}