ZOJ 3607 Lazier Salesgirl （贪心，模拟）

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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

241 2 3 41 3 6 1044 3 2 11 3 6 10

Sample Output

4.000000 2.5000001.000000 4.000000

题意：一个小姑娘烤面包卖，先给出n个客户，他们会在t[i]时间内买价值p[i]的面包，但是这个姑娘有个最大休息时间，如果每隔w时间，没有人来买面包，他就不卖了，求出使得每次平均面包价值（卖出面包总价值/卖的次数）最大的情况下，最小的间隔时间。

思路：先求出要卖出第i个面包，最小时间间隔wi[i]。在O(N)复杂度下，遍历每个顾客，当遇到一个顾客正在买第i个面包时，再想要等到第i+1个顾客需要延迟minw时，就可以计算一下当前的平均价值，并且与之前的maxave比较，取最大值，并记录minw。

注意：小女孩卖面包不是想停就能停的，需要在w间隔内没人来，才能停下，如果当前取得时间间隔minw>t[i + 1] - t[i]，这种情况下，哪怕达到的值很大，也不能停下。

代码如下：

#include<cstdio>#include<iostream>#include<cstring>#include<cmath>using namespace std;int p[1005];int t[1005];int wi[1005];int main(){int cases, n, i, j, pre;double ave, maxave, minw, sum;cin>>cases;while(cases--){cin>>n;pre = 0;sum = 0;maxave = 0;minw = 0;for(i = 0; i < n; i++){cin>>p[i];}for(i = 0; i < n; i++){//计算得要卖到当前的面包，最小间隔 cin>>t[i];wi[i] = max(t[i] - pre, wi[i - 1]);pre = t[i];}for(i = 0; i < n; i++){sum += p[i];if(i == n - 1 || wi[i + 1] > wi[i]){//如果要卖出下一个面包，需要更大的时间间隔或者已经到了最后一个，则计算一下当前的平均值 if(sum / (i + 1) > maxave){maxave = sum / (i + 1);minw = wi[i];}}}printf("%.6lf %.6lf\n", minw, maxave);}return 0;}

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