ZOJ 3607 Lazier Salesgirl (枚举)
ZOJ 3607 Lazier Salesgirl (枚举)
Lazier Salesgirl Time Limit: 2 Seconds Memory Limit: 65536 KB 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 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-th customer 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
枚举
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
int n;
int p[1005];
int t[1005];
int w;
int main()
{
int t1;
scanf("%d",&t1);
while(t1--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&p[i]);
int num1=0;
int num2=100000;
t[0]=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&t[i]);
num1=max(num1,t[i]-t[i-1]);
num2=min(num2,t[i]-t[i-1]);
}
int pos=1;
double res=0;
double ans;
for(w=num2;w<=num1;w++)
{
int sum=0;int num=0;
int time=w;
for(int i=1;i<=n;i++)
{
if(t[i]<=time)
{
sum+=p[i];
num++;
time=t[i]+w;
}
else
{
break;
}
}
double av=1.0*sum/num;
if(res<av)
{
res=av;
ans=w;
}
}
printf("%.6f %.6f\n",ans,res);
}
return 0;
}