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：

 #include<cstdio>
 #include<algorithm>
 using namespace std;
 int sum[],b[];
 int main() {
     int t;
     scanf("%d",&t);
     while(t--) {
         int n;
         scanf("%d",&n);
         int tmp;
         for(int i = ;i <= n;i++) {
             scanf("%d",&tmp);
             sum[i] = sum[i-] + tmp;
         //    printf("%d -- ",sum[i]);
         }

         for(int i = ;i <= n;i++){
             scanf("%d",&b[i]);
         }

         double ma = ;
         int maxT = ,res = ;
         for(int i = ;i <= n;i++) {
             if(b[i] - b[i - ] > maxT) {
                 maxT = b[i] - b[i - ];
             }
             while(i <= n && b[i] - b[i - ] <= maxT) i++;
             i--;
             if(1.0 * sum[i] / (i) > ma) {
                 ma = 1.0 * sum[i] / (i);
                 res = maxT;
             }
         }
         printf("%.6lf %.6lf\n",res * 1.0,ma);
     }
     return ;
 }