Geeksforgeeks Solution For " The XOR Gate "

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Problem :- Construct an N input XOR Gate. An XOR Gate returns 1 if odd number of its inputs are 1, otherwise 0.

Input:
The first line of input takes the number of test cases, T. Then T test cases follow.Each test case consists of 2 lines. The first line of each test case takes the number of inputs to the XOR Gate, N. The second line of each test case takes N space separated integers denoting the inputs to the  XOR Gate. Note that the inputs can be either 1's or 0's.


Output:
For each test case on a new line print the output of the N input XOR Gate.

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Solution :-

#include <bits/stdc++.h>
using namespace std;
int findOdd(int arr[], int n) 
{
int res = 0, i;
for (i = 0; i < n; i++)
res ^= arr[i];
return res;
}

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,i,j,sum,res=0;
        cin>>n;
        int arr[n];
        for(i=0;i<n;i++)
        cin>>arr[i];
        printf ("%d\n", findOdd(arr, n));
    }
return 0;

}

Output :-