Sicily 1046 plane spotting quick sorting

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 specified teen 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 shoshould 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-82-86-101-82-6result for run-61-71-9result for run-5

The key to this question is to understand the meaning of the question, pay attention to the first need to collect data, that is, all the time period that meets the requirements of the period, from the subscript 0 to the subscript num_ps-mp, all the qualified period values are stored in a vector of the Period type. period is a custom struct used to provide the variables required by the sorting conditions, finally, the vector is sorted by fast sorting, and the specified best period is output.

# Include <iostream> # include <vector> using namespace STD; typedef struct period {vector <int> PD; int first; int last; int length; double PQ;} period; void qsort (vector <period> &, int low, int height); int main () {INT cases; int order = 1; CIN> cases; while (Order <= cases) {int num_qs, QS, RP, MP; int enough = 0; vector <int> hold_qs; CIN> num_qs> rp> MP; vector <period> periods; For (INT I = 0; I <num_qs; I ++) {CIN> QS; hold_qs.push_back (Qs) ;}for (INT I = 0; I <= num_qs-mp; I ++) {// to the subscript is the num_ps-mp, followed by less than the minimum required length of time segment if (hold_qs [I] = 0) {// If a quarter is 0, return directly, observe the next period continue;} // search for int first = I + 1 in sequence from the end; // start with 1 instead of 0 int last = num_qs; // for the periods with the shortest length, that is, the value greater than or equal to is recorded in a vector while (last-first + 1> = Mp) {If (hold_qs [last-1]! = 0) {// double PQ = 0 from the back to the front; // This must be required; otherwise, the vector is empty period temp; periods. push_back (temp); For (Int J = first-1; j <last; j ++) {PQ + = hold_qs [J]; // calculate the number of flights per quarter. periods [enough]. PD. push_back (hold_qs [J]);} periods [enough]. first = first; periods [enough]. last = last; periods [enough]. length = last-first + 1; periods [enough]. PQ = PQ/(last-first + 1); enough ++;} Last --; // locate} // sort qsort (periods, 0, enough-1); cout <"result for Run" <order <":" <Endl; // time range required for the output. If the required number is not reached, the output is obtained. Otherwise, the output specifies the required number. If (enough <RP) {for (INT I = enough-1; I> = 0; I --) {cout <periods [I]. first <"-" <periods [I]. last <Endl ;}} else {for (INT I = enough-1; I >= enough-RP; I --) {cout <periods [I]. first <"-" <periods [I]. last <Endl ;}} order ++;} return 0;} void swap (period * a, period * B) {period temp = * A; * A = * B; * B = temp;} // fast sorting void qsort (vector <period> & periods, int low, int height) {If (low 
　　
Sicily 1046 plane spotting quick sorting