ZOJ 3607 Lazier Salesgirl (贪心)

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 nintegers 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

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

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<stdlib.h>
 4 #include<algorithm>
 5 using namespace std;
 6 const int MAXN=1000+10;
 7 int p[MAXN],t[MAXN];
 8 int main()
 9 {
10     //freopen("in.txt","r",stdin);
11     int kase;
12     scanf("%d",&kase);
13     while(kase--)
14     {
15         int n;
16         scanf("%d",&n);
17         memset(p,0,sizeof(p));
18         memset(t,0,sizeof(t));
19         for(int i=1;i<=n;i++)
20             scanf("%d",&p[i]);
21         for(int i=1;i<=n;i++)
22             scanf("%d",&t[i]);
23 
24         double maxtime=-1,time=0;
25         double sum=0,av=0;
26         for(int i=1;i<=n;i++)
27         {
28             sum+=p[i];
29             if(maxtime<t[i]-t[i-1])
30                 maxtime=t[i]-t[i-1];
31 
32             if(maxtime<t[i+1]-t[i]&&sum>av*i)
33             {
34                 av=sum/i;
35                 time=maxtime;
36             }
37 
38             if(i==n)
39             {
40                 if(av*i<sum)
41                 {
42                     av=sum/i;
43                     time=maxtime;
44                 }
45             }
46         }
47         printf("%.6lf %.6lf\n",time,av);
48     }
49     return 0;
50 }
View Code

 

posted @ 2014-08-20 22:25  Cliff Chen  阅读(342)  评论(0)  编辑  收藏  举报