CodeForce Problem 849A - Odds and Ends Solution (C++)

Problem:

You can see the question to the problem here.

Solution:

This is the simplest problem I ever had, just implement the given formulas! However, it does take me some time to write.

Code:

#include <bits/stdc++.h>

using namespace std;

main(){

int n;

int s[101];

cin>>n;

for(int i=0;i<n;i++){

    cin>>s[i];   //Numbers are stored into array

}

/*Algorithm explanation

//when n is divided by 2 remainder is not 0, result is always odd.

//Plus First element and last element.. both of them must always be odd for this Algorithm to work as mentioned in question. */

(n%2 && s[0]%2 && s[n-1]%2)?cout<<"Yes":cout<<"No";  //Another form of if else.

}