ZOJ 3607 Lazier Salesgirl (枚举)

最新推荐文章于 2020-03-03 23:02:05 发布

原创最新推荐文章于 2020-03-03 23:02:05 发布 · 575 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#ZOJ-3607 #Lazier #枚举

zoj 专栏收录该内容

28 篇文章

订阅专栏

本文探讨了一个关于懒惰售货员的算法问题，通过枚举的方法来寻找最大化面包平均售价的最佳等待时间阈值。输入包含多组测试案例，每组案例中售货员面对不同顾客的时间与价格需求，目标是确定最小的等待时间，使得售货员在不失去顾客的前提下，面包的平均售价达到最大。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 contains n 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

枚举

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>

using namespace std;
int n;
int p[1005];
int t[1005];
int w;
int main()
{
    int t1;
    scanf("%d",&t1);
    while(t1--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&p[i]);
        int num1=0;
        int num2=100000;
        t[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&t[i]);
            num1=max(num1,t[i]-t[i-1]);
            num2=min(num2,t[i]-t[i-1]);
        }
        int pos=1;
        double res=0;
        double ans;
        for(w=num2;w<=num1;w++)
        {
            int sum=0;int num=0;
            int time=w;
            for(int i=1;i<=n;i++)
            {
                if(t[i]<=time)
                {
                    sum+=p[i];
                    num++;
                    time=t[i]+w;
                }
                else
                {
                    break;
                }
            }
            double av=1.0*sum/num;
            if(res<av)
            {
                res=av;
                ans=w;
            }

        }
        printf("%.6f %.6f\n",ans,res);

    }
    return 0;
}