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Problem Statement
????
With all the different cell phone plans being offered these days, it is often difficult to pick the one that provides the best values for a particular usage pattern. In this problem, we will consider cell phone plans that provide a fixed number of minutes each month for a certain fee. If you go over the fixed number of minutes, you must pay an additional fee per minute. Additionally, some plans offer free off hour (night and weekend) calls, while other plans treat off hour calls the same as peak hour calls. If a plan offers free off hour calling, then the calls made during off hours don't use up the fixed number of minutes you get each month.  You will be given a String[], plans, each element of which represents a cell phone plan. Each plan will be formatted as "<PRICE> <MINUTES> <COST PER MINUTE> <OFF>" (quotes and angle brackets for clarity only). <PRICE> will be an integer representing the number of cents per month that one must pay for the plan. <MINUTES> will represent the number of free minutes the plan provides, per month. <COST PER MINUTE> will represent the price in cents of each additional minute in a month, beyond the first <MINUTES>. <OFF> will be either 'T' or 'F', representing whether off hour calls are free ('T') or not ('F'). Additionally, you will be given two int[]s. peakMinutes will represent the number of minutes spent talking during peak hours, while offMinutes will represent the number of minutes spent talking during off hours. Corresponding elements of peakMinutes and offMinutes will represent a single month of use.  Your task is to return the index (starting from 0) of the cheapest plan over all the months given as input. If there is a tie, return the lowest index among tied plans.
Definition
????
Class:
CellPlans
Method:
cheapest
Parameters:
String[], int[], int[]
Returns:
int
Method signature:
int cheapest(String[] plans, int[] peakMinutes, int[] offMinutes)
(be sure your method is public)
????

Constraints
-
plans will contain between 1 and 50 elements, inclusive.
-
Each element of plans will be formatted as "<PRICE> <MINUTES> <COST PER MINUTE> <OFF>".
-
<PRICE> will be between 0 and 100000, inclusive, with no extraneous leading zeros.
-
<MINUTES> will be between 0 and 50000, inclusive, with no extraneous leading zeros.
-
<COST PER MINUTE> will be between 0 and 1000, inclusive, with no extraneous leading zeros.
-
<OFF> will be either 'T' or 'F'.
-
peakMinutes will contain between 1 and 50 elements, inclusive.
-
offMinutes will contain the same number of elements as peakMinutes.
-
Each element of offMinutes and peakMinutes will be between 0 and 25000, inclusive.
Examples
0)

????
{"2999 1000 29 T",
 "5999 3000 49 F",
 "999 0 19 F"}
{1543,463,754,405,0,30}
{100,2053,1003,534,2595,3056}
Returns: 0
Plan 0 provides 1000 free minutes, and free off hour calling. For the 6 months given, we must pay the base rate (2999) each month for a total of 17994. Additionally, we used an extra 543 minutes during the first month, for which we must pay 543*29 = 15747. 15747 + 17994 = 33741.  Plan 1 would end up costing 40208, while plan 2 would end up costing 244178.
1)

????
{"10000 0 0 T","10000 0 0 F"}
{20000,25000}
{25000,20000}
Returns: 0
These plans always cost the same amount since it doesn't matter whether you get free off hour calling when the calls are already free. Since there is a tie, return the lower index.
2)

????
{"100000 0 1000 F","100000 0 10 F"}
{25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
 25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
 25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
 25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
 25000,25000,25000,25000,25000,25000}
{25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
 25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
 25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
 25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
 25000,25000,25000,25000,25000,25000}
Returns: 1
Beware of overflow.
This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

 

posted on 2005-08-16 09:29 emu 阅读(1144) 评论(1)  编辑  收藏 所属分类: google编程大赛模拟题及入围赛真题

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# emu的解法 2005-08-16 10:01 emu
送分题。主要难度是在理解题目的描述上。

public class CellPlans
{
public static void main(String[] args)
{
String [] plans = new String []{"2999 1000 29 T","5999 3000 49 F","999 0 19 F"};
int[] peakMinutes = new int[]{1543,463,754,405,0,30};
int[] offMinutes = new int[]{100,2053,1003,534,2595,3056};
int cheapest = new CellPlans().cheapest(plans,peakMinutes,offMinutes);
System.out.println("cheapest plan is :"+cheapest);

plans = new String []{"10000 0 0 T","10000 0 0 F"};
peakMinutes = new int[]{20000,25000};
offMinutes = new int[]{25000,20000};
cheapest = new CellPlans().cheapest(plans,peakMinutes,offMinutes);
System.out.println("cheapest plan is :"+cheapest);

plans = new String []{"100000 0 1000 F","100000 0 10 F"};
peakMinutes = new int[]{25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
25000,25000,25000,25000,25000,25000};
offMinutes = new int[]{25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,25000,
25000,25000,25000,25000,25000,25000};
cheapest = new CellPlans().cheapest(plans,peakMinutes,offMinutes);
System.out.println("cheapest plan is :"+cheapest);

}

public int cheapest(String[] plans, int[] peakMinutes, int[] offMinutes){
double min = count(plans[0],peakMinutes,offMinutes);
int result = 0;
for (int i=1;i<plans.length;i++){
double c = count(plans[ i ],peakMinutes,offMinutes);
if (c<min){
min=c;
result = i;
}
}
return result;
}
private double count(String plan,int[] peakMinutes,int[] offMinutes){
String[] ar = plan.split(" ");
double price = Float.parseFloat(ar[0]);
int minutes = Integer.parseInt(ar[1],10);
double costPerMin = Float.parseFloat(ar[2]);
boolean offHours = "T".equals(ar[3]);
double result = peakMinutes.length*price;
for(int i=0;i<peakMinutes.length;i++){
if (offHours){
result += costPerMin*((peakMinutes[ i ]>minutes?(peakMinutes[ i ]-minutes):0));
}else{
result += costPerMin*((peakMinutes[ i ]+offMinutes[ i ])>minutes?(peakMinutes[ i ]+offMinutes[ i ]-minutes):0);
}
}
return result;
}

}
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