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

 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