# XOR Matrix

### Problem Statement :

```Consider a zero-indexed matrix with  rows and  columns, where each row is filled gradually. Given the first row of the matrix, you can generate the elements in the subsequent rows using the following formula:

Each row is generated one by one, from the second row through the last row. Given the first row of the matrix, find and print the elements of the last row as a single line of space-separated integers.

Note: The  operator denotes bitwise XOR.

Input Format

The first line contains two space-separated integers denoting the respective values of  (the number of columns in the matrix) and  (the number of rows in the matrix).
The second line contains  space-separated integers denoting the respective values of the elements in the matrix's first row.

Output Format

Print  space-separated integers denoting the respective values of the elements in the last row of the matrix.```

### Solution :

```                            ```Solution in C :

In  C  :

#include <stdio.h>
#include <stdlib.h>

int main(int argc, const char * argv[]) {
long n;
scanf("%li", &n);
long long m;
scanf("%lli", &m);
m -= 1;
int i;
long *a = (long *)malloc(n * sizeof(long));
long *b = (long *)malloc(n * sizeof(long));
long *t;

for(i=0; i<n; i++) {
scanf("%li", a+i);
}

long shift = 1;
while (m > 0) {
if (m & 1) {
for(i=0; i<n; i++) {
if (i+shift < n) {
b[i] = a[i] ^ a[i+shift];
} else {
b[i] = a[i] ^ a[i+shift-n];
}
}
t = a;
a = b;
b = t;
}
shift <<= 1;
if (shift >= n) {
shift -= n;
}
m >>= 1;
}

printf("%li", a[0]);
for(i=1; i<n; i++) {
printf(" %li", a[i]);
}
printf("\n");
return 0;
}```
```

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

In  C++ :

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <ctype.h>
#include <deque>
#include <queue>
#include <cstring>
#include <set>
#include <list>
#include <map>
#include <random>
#include <unordered_map>
#include <stdio.h>

using namespace std;

typedef long long ll;
typedef std::vector<int> vi;
typedef std::vector<bool> vb;
typedef std::vector<string> vs;
typedef std::vector<double> vd;
typedef std::vector<long long> vll;
typedef std::vector<std::vector<int> > vvi;
typedef vector<vvi> vvvi;
typedef vector<vll> vvll;
typedef std::vector<std::pair<int, int> > vpi;
typedef vector<vpi> vvpi;
typedef std::pair<int, int> pi;
typedef std::pair<ll, ll> pll;
typedef std::vector<pll> vpll;

const long long mod = 1000000007;

#define all(c) (c).begin(),(c).end()
#define sz(c) (int)(c).size()
#define forn(i, a, b) for(int i = a; i < b; i++)

#define pb push_back
#define mp make_pair

int main()
{

int n;
ll m;
scanf("%d %lld", &n, &m);
m--;
vi a(n);
vll d2(1,1);
forn(i,0,60) d2.pb(d2.back()*2);
forn(i,0,n) scanf("%d", &a[i]);
for(int bit = 60; bit>=0; bit--) {
if(m>=d2[bit]) {
vi b(n);
forn(i,0,n) {

b[i] = a[i]^a[((ll)i+d2[bit])%n];

}
a=std::move(b);
m-=d2[bit];
}
}
forn(i,0,n) printf("%d ", a[i]);

}```
```

```                        ```Solution in Java :

In  Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
long m = sc.nextLong()-1;
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = sc.nextInt();
int shift = 1;
while (m > 0) {
if (m%2==1) {
int[] newa = new int[n];
for (int i = 0; i < n; i++) {
newa[i] = a[i]^a[(i+shift)%n];
}
a = newa;
}
m /= 2;
shift *= 2;
shift %= n;
}
StringBuilder ans = new StringBuilder();
ans.append(a[0]);
for (int i = 1; i < n; i++) {
ans.append(" "+a[i]);
}
System.out.println(ans);
}
}```
```

```                        ```Solution in Python :

In  Python3 :

n, m = map(int, input().split())
ar = tuple(map(int, input().split()))
i = 1
m -= 1
while m:
if m & 1:
j = i % n
ar = tuple(ar[pos] ^ ar[(pos + j) % n] for pos in range(n))
m >>= 1
i <<= 1
print(*ar)```
```

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