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# War of XORs CodeChef Solution

## War of XORs CodeChef Solution in C++14

``````#include <bits/stdc++.h>
using namespace std;
#define ll long long int

int main()
{
int t;
cin >> t;
while (t--)
{
int n;
cin >> n;
int i, arr[n], freq[1000010] = {0};
for (int i = 0; i < n; i++)
{
cin >> arr[i];
freq[arr[i]]++;
}
ll odd = 0, even = 0, ans = 0;
for (int i = 0; i < n; i++)
{
if (arr[i] & 1)
odd++;
else
even++;
}

for (int i = 0; i < n; i++)
{
if (arr[i] & 1)
ans += odd;
else
ans += even;
ans -= freq[arr[i] ^ 2];
ans -= freq[arr[i]];
}
cout << (ans >> 1) << endl;
}

return 0;
}``````

## War of XORs CodeChef Solution in PYTH 3

``````# cook your dish here
from collections import defaultdict
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
d=defaultdict(lambda:0)
even=odd=0
count=0
for i in l:
d[i]+=1
if i&1:odd+=1
else:even+=1
for i in l:
if(i%2 == 0):
count = count + even
else:
count = count + odd
if(i ^ 2 in d):
count = count - d[i^2]
count = count - d[i]
print(int(count/2))``````

## War of XORs CodeChef Solution in C

``````#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define int long long int
#define li long int
#define inf 1000010
#define mod 998244353

int Floor(int a, int b)
{
if(a%b == 0)
{
return (a/b);
}
if(a>=0&&b>=0 || a<=0&&b<=0)
{
return a/b;
}
else
{
return (a/b - 1);
}
}

int max(int a, int b)
{
int m = a;
if(b>a)
m = b;
return m;
}

int min(int a, int b)
{
int m = a;
if(b<a)
m = b;
return m;
}

// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(int arr[], int l, int m, int r)
{
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;

/* create temp arrays */
int L[n1], R[n2];

/* Copy data to temp arrays L[] and R[] */
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1 + j];

/* Merge the temp arrays back into arr[l..r]*/
i = 0; // Initial index of first subarray
j = 0; // Initial index of second subarray
k = l; // Initial index of merged subarray
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
}
else {
arr[k] = R[j];
j++;
}
k++;
}

/* Copy the remaining elements of L[], if there
are any */
while (i < n1) {
arr[k] = L[i];
i++;
k++;
}

/* Copy the remaining elements of R[], if there
are any */
while (j < n2) {
arr[k] = R[j];
j++;
k++;
}
}

/* l is for left index and r is right index of the
sub-array of arr to be sorted */
void mergeSort(int arr[], int l, int r)
{
if (l < r) {
// Same as (l+r)/2, but avoids overflow for
// large l and h
int m = l + (r - l) / 2;

// Sort first and second halves
mergeSort(arr, l, m);
mergeSort(arr, m + 1, r);

merge(arr, l, m, r);
}
}

int H[inf];

void init()
{
int i;
for ( i = 0; i < inf; i++)
{
H[i] = 0;
}
return;
}

void close(int D[], int lend)
{
for (int i = 0; i < lend; i++)
{
H[D[i]] = 0;
}
return;
}

void program()
{
int n;
scanf("%lld", &n);
int A[n];
int D[n];
int i;
for ( i = 0; i < n; i++)
{
scanf("%lld", &A[i]);
}
//only even numbers >=4  can be sum of same parity prime numbers
int j = 0;
for ( i = 0; i < n; i++)
{
if(H[A[i]] == 0)
{
D[j] = A[i];
H[A[i]]++;
j++;
}
else
{
H[A[i]]++;
}
}
int len_d = j;
int even = 0, odd = 0;
for ( i = 0; i < n; i++)
{
if(A[i]%2 == 0)
{
even++;
}
else
{
odd++;
}
}
/*for ( i = 0; i < len_d; i++)
{
printf("%d %d\n", D[i], H[D[i]]);
}
printf("%d %d\n", even, odd);*/
int ans = (even)*(even-1) + (odd)*(odd-1);
for ( i = 0; i < len_d; i++)
{
int b = (D[i])^2;
ans -= H[b]*H[D[i]];
ans -= (H[D[i]])*(H[D[i]]-1);
}
printf("%lld\n", ans/2);
close(D, len_d);
}

int main(void)
{
init();
int T;
scanf("%lld", &T);
//printf("%d\n", T);
// T = 1;
for (int i = 0; i < T; i++)
{
program();
}
return 0;
}
``````

## War of XORs CodeChef Solution in JAVA

``````/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
int t = sc.nextInt();
while (t-- > 0){
int n = sc.nextInt();
long f[] = new long[1000011];
long one = 0;
long res = 0;
for (int i = 0 ; i < n ; i++){
long temp = sc.nextInt();
res = res - f[(int)temp] - f[(int)temp^2];
if ((temp & 1l) == 1l) res = res + one++;
else res = res + ((long)i - one);
f[(int)temp]++;
}
System.out.println(res);
}
}

StringTokenizer st;

try {
}catch (Exception e) {
System.out.println(e.toString());
}
}

String next(){
while (st == null || !st.hasMoreElements()){
try{
}catch (IOException  e){
e.printStackTrace();
}
}
return st.nextToken();
}

int nextInt(){
return Integer.parseInt(next());
}

long nextLong(){
return Long.parseLong(next());
}

double nextDouble(){
return Double.parseDouble(next());
}

String nextLine(){
String str = "";
try{
}catch (IOException e){
e.printStackTrace();
}
return str;
}
}
}``````

## War of XORs CodeChef Solution in PYPY 3

``````from collections import defaultdict
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
d=defaultdict(lambda:0)
even=odd=0
count=0
for i in l:
d[i]+=1
if i&1:odd+=1
else:even+=1
for i in l:
if(i%2 == 0):
count = count + even
else:
count = count + odd
if(i ^ 2 in d):
count = count - d[i^2]
count = count - d[i]
print(int(count/2))``````

## War of XORs CodeChef Solution in PYTH

``````MX = 10**6
t = int(raw_input())
for i in range(t):
N = int(raw_input())
st = raw_input().split()
F = [0 for x in range(MX+1)]
ne = 0
for x in st:
n = int(x)
if n%2 == 0:
ne += 1
# endif
F[n] += 1
# endfor x
no = N-ne
tot = no*(no-1)/2 + ne*(ne-1)/2
for n in range(1,MX+1):
m = F[n]
if m > 0:
tot -= m*(m-1)/2
if n&2 > 0:
k = F[n-2]
tot -= m*k
# endif
# endif
# endfor x
print tot
# endfor i
``````

## War of XORs CodeChef Solution in C#

``````using System;
using System.Collections.Generic;

namespace ChallengeSolver.CodeChef
{
/// <summary>
/// https://www.codechef.com/SEPT18B/problems/XORIER
/// </summary>
static class Xorier
{
public static void Main()
{
for (int t = 0; t < testCount; t++)
{
long evenCount = 0;
long oddCount = 0;
Dictionary<int, long> evenSet = new Dictionary<int, long>();
Dictionary<int, long> oddSet = new Dictionary<int, long>();
for (int i = 0; i < rawInput.Length; i++)
{
int value = int.Parse(rawInput[i]);
if ((value & 1) == 0)
{
oddCount++;
if (!evenSet.ContainsKey(value))
{
}
else
{
evenSet[value]++;
}
}
else
{
evenCount++;
if (!oddSet.ContainsKey(value))
{
}
else
{
oddSet[value]++;
}
}
}

long result = evenCount * (evenCount - 1) / 2 + oddCount * (oddCount - 1) / 2;
long componentCount = 0;
foreach (var i in evenSet)
{
result -= i.Value * (i.Value - 1) / 2;
int complement = i.Key ^ 2;
if (evenSet.ContainsKey(complement))
{
componentCount += (i.Value * evenSet[complement]);
}
}
result -= componentCount / 2;
componentCount = 0;
foreach (var i in oddSet)
{
result -= i.Value * (i.Value - 1) / 2;
int complement = i.Key ^ 2;
if (oddSet.ContainsKey(complement))
{
componentCount += (i.Value * oddSet[complement]);
}
}
result -= componentCount / 2;
Console.WriteLine(result);
}
}
}
}``````
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