Given number n find the sum of the even positive numbers before it-CodePudding

here given the number 'n' we have find the sum of the even positive numbers before it. For that I tried to put a while loop in a for loop,but looks like that wont work

#include<iostream>

using namespace std;

int main() {

    int n;
    cin>>n;

    int sum=0;
    for(int counter=1;counter<=n;counter  )
        while(counter%2==0) {
            sum=sum counter;
        }    
    cout<<sum<<endl;

    return 0;
}

CodePudding user response：

First off, I'd highly recommend incorporating some sort of UI, as it'll help you keep track of what's going on as your programs grow in complexity:

    cout << "Enter an integer, and I will calculate the sum of all its  preceding, positive integers: ";

And you'd need to swap that "while" for an "if". Reason being, you're adding to the sum, IF the counter's value is even:

for(int counter=1; counter<=n; counter  ){
   if(counter%2==0){
    sum=sum counter;
   }
}

Hope that helps!

CodePudding user response：

You probably want:

for(int counter=1;counter<=n;counter  )
    if(counter%2==0) // this is an if condition, not another loop
    {
        sum=sum counter;
    }

CodePudding user response：

Rather than iterating over numbers, we can solve this in O(1) time complexity.
Notice that sum of even numbers before a number N, forms an Arithmetic progression with 2 as first term, common difference 2 and number of terms ⌊N/2⌋

Sum = 2   4   6 ... N
Sum = 2(1   2   3 ... N/2)  
Sum = 2((n)*(n 1)/2)   
Sum = n*(n 1)

where n = ⌊N/2⌋

#include <iostream>
using namespace std; 

int main() {
    int N
    cin>>N;
    N = max(0, N);
    
    long n = N/2;
    
    long sum = n*(n 1);
    
    cout<<sum<<endl;
    
    return 0;
}