zjuoj 3607 Lazier Salesgirl

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

Lazier Salesgirl

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

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 }