zjuoj 3607 Lazier Salesgirl

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3607

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

 

分析：

求当睡觉时间最短，卖出的面包平均价格最高时的Ｗ和平均值，注意：当他睡着的时候就不再醒啦，，，

 

 

ＡＣ代码：

 1 #include<cstdio>

 2 #include<algorithm>

 3 using namespace std;

 4 int sum[1005],b[1005];

 5 int main() {

 6     int t;

 7     scanf("%d",&t);

 8     while(t--) {

 9         int n;

10         scanf("%d",&n);

11         int tmp;

12         for(int i = 1;i <= n;i++) {

13             scanf("%d",&tmp);

14             sum[i] = sum[i-1] + tmp;

15         //    printf("%d -- ",sum[i]);

16         }

17         

18         for(int i = 1;i <= n;i++){

19             scanf("%d",&b[i]);

20         }

21         

22         double ma = 0;

23         int maxT = 0,res = 0;

24         for(int i = 1;i <= n;i++) {

25             if(b[i] - b[i - 1] > maxT) {

26                 maxT = b[i] - b[i - 1];

27             }

28             while(i <= n && b[i] - b[i - 1] <= maxT) i++;

29             i--;

30             if(1.0 * sum[i] / (i) > ma) {

31                 ma = 1.0 * sum[i] / (i);

32                 res = maxT;

33             }

34         }

35         printf("%.6lf %.6lf\n",res * 1.0,ma);

36     }

37     return 0;

38 }

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