# Sansa and XOR

### Problem Statement :

```Sansa has an array. She wants to find the value obtained by XOR-ing the contiguous subarrays, followed by XOR-ing the values thus obtained. Determine this value.

Example

Subarray	Operation	Result
3		None		3
4		None		4
5		None		5
3,4		3 XOR 4		7
4,5		4 XOR 5		1
3,4,5		3 XOR 4 XOR 5	2
Now we take the resultant values and XOR them together:

. Return .

Function Description

Complete the sansaXor function in the editor below.

sansaXor has the following parameter(s):

int arr[n]: an array of integers
Returns

int: the result of calculations

Input Format

The first line contains an integer , the number of the test cases.

Each of the next  pairs of lines is as follows:
- The first line of each test case contains an integer , the number of elements in .
- The second line of each test case contains  space-separated integers .```

### Solution :

```                            ```Solution in C :

In  C :

#include <stdio.h>
#include <stdlib.h>
#define MOD 1000000007
#define s(k) scanf("%d",&k);
#define sll(k) scanf("%lld",&k);
#define p(k) printf("%d\n",k);
#define pll(k) printf("%lld\n",k);
#define f(i,N) for(i=0;i<N;i++)
#define f1(i,N) for(i=0;i<=N;i++)
#define f2(i,N) for(i=1;i<=N;i++)
typedef long long ll;
int main(){
int t,n,k,sum,i;
s(t)
while(t--){
s(n)
if(!(n%2)){
printf("0\n");
f(i,n)
s(k)
} else {
sum=0;
f(i,n){
s(k)
if(!(i%2))
sum^=k;
}
p(sum);
}
}
return 0;
}```
```

```                        ```Solution in C++ :

In  C++  :

#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <limits>
#include <cstring>
#include <string>
using namespace std;

#define pairii pair<int, int>
#define llong long long
#define pb push_back
#define sortall(x) sort((x).begin(), (x).end())
#define INFI  numeric_limits<int>::max()
#define INFLL numeric_limits<llong>::max()
#define INFD  numeric_limits<double>::max()
#define FOR(i,s,n) for (int (i) = (s); (i) < (n); (i)++)
#define FORZ(i,n) FOR((i),0,(n))

void solve() {
int n;
scanf("%d",&n);
int res = 0;
FORZ(i,n) {
int x;
scanf("%d",&x);
if ((i+1)%2 && (n-i)%2) {
res ^= x;
}
}
printf("%d\n",res);
}

int main() {
#ifdef DEBUG
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
int t;
scanf("%d",&t);
FORZ(i,t) solve();
return 0;
}```
```

```                        ```Solution in Java :

In  Java :

import java.io.IOException;
import java.util.HashMap;
import java.util.Map;

public class Solution {
public static void main(String[] args) throws IOException {
int num = Integer.parseInt(line);

for (int i = 0; i < num; i++) {
sansaXor(ns, Integer.valueOf(count));
}

}

public static void sansaXor(String[] strs, int count){
Map<Integer,Integer> map = new HashMap<Integer,Integer>();
long result = 0;
for(int i = 0; i < count; i++){
int numCount = (i + 1)*(count - i);
int tmp = Integer.valueOf(strs[i]);
if(map.containsKey(tmp)){
map.put(tmp, numCount+map.get(tmp));
}else{
map.put(tmp, numCount);
}
}

for(int k: map.keySet()){
int value = map.get(k);
if(value%2!=0){
result = result^k;
}
}

System.out.println(result);
}
}```
```

```                        ```Solution in Python :

In  Python3 :

for _ in range(int(input())):
N = int(input())
A = tuple(map(int,input().split()))
X = 0
for i, x in enumerate(A):
if ((i+1) * (N-i)) % 2 == 1:
X ^= x
print(X)```
```

## Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink

Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below

## Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e

## Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

## Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing