Thursday, 2 February 2017

Geeksforgeeks Solution For " The XOR Gate "

GeeksforGeeks Solution For School Domain .Below You Can Find The Solution Of Basic ,Easy ,Medium ,Hard .You Can Also Direct Submit Your Solution to Geeksforgeeks Same Problem .You Need to login then you can submit you answers 

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.

Submit Your Solution Click Here 


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 :-

Geeksforgeeks Solution For " The XOR Gate "

No comments:
Write comments

Recommended Posts × +