ZOJ 3607 Lazier Salesgirl（贪心啊 ）

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4710

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 n integers 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

241 2 3 41 3 6 1044 3 2 11 3 6 10

Sample Output

4.000000 2.5000001.000000 4.000000

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Author: WU, Zejun
Contest: The 9th Zhejiang Provincial Collegiate Programming Contest

代码如下：

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;int max(int a, int b){if(a > b){return a;}return b;}int main(){int T;int n;int p[1017], t[1017];scanf("%d",&T);while(T--){int i, j;scanf("%d",&n);for(i = 0;  i < n; i++){scanf("%d",&p[i]);}for(i = 0;  i < n; i++){scanf("%d",&t[i]);}int c[1017];c[0] = t[0];for(i = 1; i < n; i++){c[i] = max(c[i-1], t[i]-t[i-1]);}double sum = 0;double w, ave = 0;for(i = 0; i < n; i++){double tt = c[i];sum = 0;for(j = 0; j < n; j++){if(tt >= c[j]){sum+=p[j];}else{break;}}if(sum/j > ave){ave = sum/j;w = c[i];}}printf("%.6lf %.6lf\n",w,ave);}return 0;}