ZOJ 3607 Lazier Salesgirl (贪心)

Lazier Salesgirl

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

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

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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

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
const int MAXN=1000+10;
int p[MAXN],t[MAXN];
int main()
{
    //freopen("in.txt","r",stdin);
    int kase;
    scanf("%d",&kase);
    while(kase--)
    {
        int n;
        scanf("%d",&n);
        memset(p,0,sizeof(p));
        memset(t,0,sizeof(t));
        for(int i=1;i<=n;i++)
            scanf("%d",&p[i]);
        for(int i=1;i<=n;i++)
            scanf("%d",&t[i]);

        double maxtime=-1,time=0;
        double sum=0,av=0;
        for(int i=1;i<=n;i++)
        {
            sum+=p[i];
            if(maxtime<t[i]-t[i-1])
                maxtime=t[i]-t[i-1];

            if(maxtime<t[i+1]-t[i]&&sum>av*i)
            {
                av=sum/i;
                time=maxtime;
            }

            if(i==n)
            {
                if(av*i<sum)
                {
                    av=sum/i;
                    time=maxtime;
                }
            }
        }
        printf("%.6lf %.6lf
",time,av);
    }
    return 0;
}