zoj 3607 Lazier Salesgirl

Lazier SalesgirlTime Limit: 2 Seconds      Memory Limit: 65536 KBKochiya 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 pi.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-thcustomer come after ti 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 ≤ pi ≤10000. The third line contains n integers 1 ≤ ti≤ 100000. The customers are given in the non-decreasing order of ti.

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

#include <bits/stdc++.h>using namespace std;const int N = 1e4 + 10;int t, n;double p, pp, v, vv;struct xx{double p, v;int n;}a;bool cmp(xx a, xx b){if(a.p == b.p) return a.v < b.v;return a.p > b.p;}int main(){scanf("%d", &t);while(t--){scanf("%d", &n);double sum = 0, maxi = 0;int ii = 0;for(int i = 1; i <= n; i++){scanf("%lf", &p[i]);p[i] += p[i-1];//printf("%.3f ", p[i]);}//printf("\n");for(int i = 1; i <= n; i++){scanf("%lf", &v[i]);double tmp = v[i]-v[i-1];vv[i] = max(vv[i-1], tmp);//printf("%.3f ", vv[i]);}//printf("\n");v[n+1] = 0x3f3f3f3f;int x = 0;for(int i = 1; i <= n; i++){if(vv[i] < v[i+1]- v[i]){a[x].v = vv[i];a[x].p = p[i]/i*1.0;x++;}}sort(a, a+x, cmp);printf("%.6f %.6f\n", a[0].v, a[0].p);}}