Hint: Missed newline ? while trying to solve the below quiz [closed] - c++

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Problem
This is similar to the previous problem, however there will be multiple integers in the input. You have to write a computer program to read each integer and print Even if the integer is divisible by 2, else print Odd. To help further, the number of integers (T) to read will be the first input to the computer program.
Input Format:
First line of input contains count of integers: T. T>=1
After that, each line contains the integer N.
Sample Input:
2
4
5
Sample Output:
Even
Odd
Note: There should be a newline after each output. Otherwise you might end up printing EvenOdd here, which will result in wrong answer.
#include <iostream>
using namespace std;
int main()
{
int a[3];
cin>>a[0];
cin>>a[1];
cin>>a[2];
for(int i =0; i <sizeof(a)/sizeof(int); i++)
{
if (a[i]%2 == 0) cout<<"Even"<<endl;
else cout<<"Odd"<<endl;
}
return 0;
}

Answer 1

You don't need neither array nor vectors for this:
int main() {
int n;
cin >> n;
while(n--) {
int number;
cin >> number;
// do something with number
}
return 0;
}
Note that this structure is the most common to solve this kind of problems

Answer 2

your problem will be solved if you change
for(int i =0; i <sizeof(a)/sizeof(int); i++)
to
for(int i =1; i <sizeof(a)/sizeof(int); i++)
But, how'd you know the number of elemnts for the array beforehand?
To correct this, you don't declare the array statically. Take the first input, and allocate memory for that many ints.

Answer 3

#include <iostream>
using namespace std;
int main()
{
int a[3];
cin>>a[0];
cin>>a[1];
cin>>a[2];
for(int i =0; i <sizeof(a)/sizeof(int); i++)
{
if (a[i]%2 == 0)
cout<<"Even"<<endl;
else
cout<<"Odd"<<endl;
}
return 0;
}
this code which u posted is working perfectly bro :)