Negative Numbers

#1

(did you account for negative numbers, for example?). When ordering tasks that include negative values, the greedy algorithm seems to work fine. What is the problem with negative numbers?

#2

Hey, negative numbers in what terms, we are given time so they cannot be negative neither the interval be negative work cannot end before it has started.

I am willing to understand where are u willing to introduce the concept of negative numbers in Activity selection

#3

I agree that there can be no negative numbers, precisely when uploading my code, the observation made by the feedback system for the second test case is if I am taking into account negative numbers and, like you, I ask myself refers. That is my question.

#4

could u elaborate i could not get ur point… like what is ur doubt
tell me the instance when u are considering -ve number i/p

#5

This is the system message after uploading my code. He suggests taking negative numbers into account. I think it is not correct.

Error%20message
Error%20message1126×484 18 KB

#6

send me the code in coding blocks ide
there would be some error in the code
i`ll help u out

#7
#include <iostream

#include <algorithm

using namespace std;

int t,n;
pair<int,int> A[105];

bool comparador(pair<int,int> a, pair<int,int> b){
if (a.second < b.second) return true;
return false;
}

int main() {
cin >> t;
// cout << t << endl;

for (int i=0; i< t; i++) { // para cada tarea a resolver
	cin >> n; // leer número de actividades

// cout << n << endl;
for (int j=0; j< n;j++) {
cin >> A[j].first >> A[j].second;
// cout << A[j].first << " " << A[j].second << endl;
}

	sort(A,A+n,comparador);

    int cuenta=0;
	int previa =0;
	for (int j=0; j< n;j++) {
        
        if ( j == 0){ 
		    //cout << A[j].first << " " << A[j].second << endl;
            previa = j;
            cuenta++;
        }
        else {
            if ( A[previa].second <= A[j].first) {
                //cout << A[j].first << " " << A[j].second << endl;
                previa = j;
                cuenta++;
            }
        }
	}
    cout << cuenta;

}
return 0;

}
#8

just a small mistake u havent added a new line character
cout << cuenta << endl;

will give u the correct answer

#9

You’re right, I didn’t see it. Line break is needed to separate cases.

Thanks for yout help.

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