sicily1046

1046. Plane Spotting

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Craig is fond of planes. Making photographs of planes forms a major part of his daily life. Since he tries to stimulate his social life, and since it’s quite a drive from his home to the airport, Craig tries to be very efficient by investigating what the optimal times are for his plane spotting. Together with some friends he has collected statistics of the number of passing planes in consecutive periods of fifteen minutes (which for obvious reasons we shall call ‘quarters’). In order to plan his trips as efficiently as possible, he is interested in the average number of planes over a certain time period. This way he will get the best return for the time invested. Furthermore, in order to plan his trips with his other activities, he wants to have a list of possible time periods to choose from. These time periods must be ordered such that the most preferable time period is at the top, followed by the next preferable time period, etc. etc. The following rules define which is the order between time periods:

1. A period has to consist of at least a certain number of quarters, since Craig will not drive three hours to be there for just one measly quarter. 
2. A period P1 is better than another period P2 if: 
* the number of planes per quarter in P1 is higher than in P2; 
* the numbers are equal but P1 is a longer period (more quarters); 
* the numbers are equal and they are equally long, but period P1 ends earlier.

Now Craig is not a clever programmer, so he needs someone who will write the good stuff: that means you. So, given input consisting of the number of planes per quarter and the requested number of periods, you will calculate the requested list of optimal periods. If not enough time periods exist which meet requirement 1, you should give only the allowed time periods.

Input

The input starts with a line containing the number of runs N. Next follows two lines for each run. The first line contains three numbers: the number of quarters (1–300), the number of requested best periods (1–100) and the minimum number of quarters Craig wants to spend spotting planes (1–300). The sec-nod line contains one number per quarter, describing for each quarter the observed number of planes. The airport can handle a maximum of 200 planes per quarter.

Output

The output contains the following results for every run: 

* A line containing the text “Result for run <N>:” where <N> is the index of the run. 

* One line for every requested period: “<F>-<L>” where <F> is first quarter and <L> is the last quarter of the period. The numbering of quarters starts at 1. The output must be ordered such that the most preferable period is at the top.

Sample Input

310 5 51 5 0 2 1 4 2 5 0 2 10 3 510 3 1 4 2 6 3 0 8 0 5 5 51 2 3 4 5

Sample Output

Result for run 1:4-82-86-101-82-6Result for run 2:1-61-71-9Result for run 3:1-5

这道题目并不难，主要是题目的意思要看清。

给出的Input中的每个测试数据中的第一行的三个数字分别是：给出的quarters总数；输出的最大period数；每个period必须包含的最小quarters数。

数据结构：

STL的vector

struct

采用vector的原因是这道题要排序，而且结构体的数目未知，可能比较大。如果单纯用数组来做就会面临排序的困难。

要点：

对vector排序

用algorithm 库中的sort。要自己重写comp函数。

/* * main.cpp * *  Created on: 2014年9月23日 *      Author: xiangxiyun */#include <iostream>#include <vector>#include <algorithm>using namespace std;struct period {double begin, end;double planes;double quarters;};bool Comp(const period &a, const period &b) {if (a.planes > b.planes)return true;else if (a.planes == b.planes && a.quarters > b.quarters)return true;else if (a.planes == b.planes && a.quarters == b.quarters && a.end < b.end)return true;return false;}int main() {int n;cin >> n;for (int k = 1; k <= n; k++) {int sum, maxi, mini;cin >> sum >> maxi >> mini;vector<period> lis;period hold;int a[sum + 1];for (int i = 1; i <= sum; i++)cin >> a[i];for (int i = 1; i <= sum - mini + 1; i++) {int plane = 0;for (int j = i; j < i + mini - 1; j++)plane += a[j];for (int j = i + mini - 1; j <= sum; j++) {plane += a[j];hold.begin = i;hold.end = j;hold.quarters = j - i + 1;hold.planes = plane / hold.quarters;lis.push_back(hold);}}sort(lis.begin(), lis.end(), Comp);/*for (int i = 0; i < lis.size(); i++) {cout << (lis[i]).begin << "-" << lis[i].end << endl;cout << (lis[i]).quarters << "   " << lis[i].planes<< endl;}*/cout << "Result for run " << k << ":" << endl;if (lis.size() < maxi) {for (int i = 0; i < lis.size(); i++) {cout << (lis[i]).begin << "-" << lis[i].end << endl;}} else {for (int i = 0; i < maxi; i++) {cout << (lis[i]).begin << "-" << lis[i].end << endl;}}}return 0;}