way to learn!!

Question

i was solving problem on codeforce and faced a new input method to me (sequence of numbers separated by spaces) and failed to solve it, i searched for the solution and found it::
#include <iostream>
#include <algorithm>
using namespace std;

int main(){
int n, k; cin >> n >> k;
int *scores = new int[n];
for(int u = 0; u < n; ++u){cin >> scores[u];}
sort(scores, scores + n);
int min = scores[n - k];
int promoted = 0;
while(promoted < n && scores[n-1 - promoted] > 0 && scores[n-1 - promoted] >= min){++promoted;}
cout << promoted << endl;
return 0;
}

*** i know that my weak understanding is the cause but can any one describe the action of this code to me??

Answer 1

"int n, k; cin >> n >> k;"

This part can really be split up in for different lines to make it slightly easier to read:

int n;
int k;
cin >> n;
cin >> k;

This defines to integer variables and then asks the user to input to integers, storing the first in "n" and the second in "k".

"int *scores = new int[n];"

This defines an integer pointer that dynamically allocates memory via the "new" keyword for an integer array of size "n" (the first user input integer).

"for(int u = 0; u < n; ++u) {
cin >> scores[u];
}"

This for loop iterates through each spot in scores (the integer array) and asks the user to enter a number to put in each spot.

"sort(scores, scores + n);"

sort is a method included in the artay class the takes the starting and ending index of the array you want sorted, and sorts them from smallest to largest.

"int min = scores[n - k];"

defines an integer "min" as the value at index "n-k" in the scores array.

"int promoted = 0;"

Defines a new integer "promoted" with a value of 0.

Answer 2

"while(promoted < n && scores[n - 1 - promoted] > 0 && scores[n - 1 - promoted] >= min) {
++promoted;
}

This piece of code is essentially starting at the back of the array (scores[n - 1 - (0)]) and increasing "promoted" by 1 until it reaches a value that is either less than 0 (scores[n - 1 - promoted] > 0) or less than or equal to min (scores[n - 1 - promoted] >= min).

"cout << promoted << endl;"

prints out the value of promoted.

"return 0;"

ends the main method.

I apologize for going literally line by line, but i was not sure what you do and do not already know.