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 nintegers 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
Author: WU, Zejun
Contest: The 9th Zhejiang Provincial Collegiate Programming Contest
题意：
T个情况，每个情况有n个客人，第i个客人可赚p[i]元钱，第i个客人t[i]时间来，卖东西的女孩很懒，如果w时间内没人来就睡觉，后面来的客人就会刚来就走，求赚钱的平均值最大同时输出最小w.
这题采用贪心，如果t[i+1]-t[i]<maxtime时，那么后面一个客人能买到面包，如果t[i+1]-t[i]>maxtime那么就先把当前的情况记录下来，再往后去找
1 #include<cstdio>
2 #include<cstring>
3 #include<stdlib.h>
4 #include<algorithm>
5 using namespace std;
6 const int MAXN=1000+10;
7 int p[MAXN],t[MAXN];
8 int main()
9 {
10 //freopen("in.txt","r",stdin);
11 int kase;
12 scanf("%d",&kase);
13 while(kase--)
14 {
15 int n;
16 scanf("%d",&n);
17 memset(p,0,sizeof(p));
18 memset(t,0,sizeof(t));
19 for(int i=1;i<=n;i++)
20 scanf("%d",&p[i]);
21 for(int i=1;i<=n;i++)
22 scanf("%d",&t[i]);
23
24 double maxtime=-1,time=0;
25 double sum=0,av=0;
26 for(int i=1;i<=n;i++)
27 {
28 sum+=p[i];
29 if(maxtime<t[i]-t[i-1])
30 maxtime=t[i]-t[i-1];
31
32 if(maxtime<t[i+1]-t[i]&&sum>av*i)
33 {
34 av=sum/i;
35 time=maxtime;
36 }
37
38 if(i==n)
39 {
40 if(av*i<sum)
41 {
42 av=sum/i;
43 time=maxtime;
44 }
45 }
46 }
47 printf("%.6lf %.6lf\n",time,av);
48 }
49 return 0;
50 }
ZOJ 3607 Lazier Salesgirl (贪心),布布扣,bubuko.com
时间： 2024-10-12 03:26:38