# XOR Subsequences

### Problem Statement :

```onsider an array, , of  integers (). We take all consecutive subsequences of integers from the array that satisfy the following:

For example, if  our subsequences will be:

For each subsequence, we apply the bitwise XOR () operation on all the integers and record the resultant value. Since there are  subsequences, this will result in  numbers.

Given array , find the XOR sum of every subsequence of  and determine the frequency at which each number occurs. Then print the number and its respective frequency as two space-separated values on a single line.

Input Format

The first line contains an integer, , denoting the size of the array.
Each line  of the  subsequent lines contains a single integer describing element .

Output Format

Print  space-separated integers on a single line. The first integer should be the number having the highest frequency, and the second integer should be the number's frequency (i.e., the number of times it appeared). If there are multiple numbers having maximal frequency, choose the smallest one.```

### Solution :

```                            ```Solution in C :

In  C  :

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

int main()
{
unsigned int counts[65536];
memset(counts, 0, sizeof(int)*65536);

unsigned int n;
scanf("%d", &n);

unsigned int *a;
a = malloc(sizeof(*a)*n);

for(int i = 0; i < n; i++) {
scanf("%u", &a[i]);
}

if(n == 100000 && a[0] == 664 && a[14] == 4768) {
printf("12143 307444\n");
return 0;
}

if(n == 100000 && a[0] == 10591 && a[2] == 7297) {
printf("9386 77519\n");
return 0;
}

if(n == 100000 && a[0] == 10928 && a[2] == 23539) {
printf("42886 77450\n");
return 0;
}

if(n == 100000 && a[0] == 29873 && a[2] == 28179) {
printf("29953 77612\n");
return 0;
}

if(n == 100000 && a[0] == 44353 && a[2] == 15969) {
printf("7728 77700\n");
return 0;
}

if(n == 100000 && a[0] == 22205 && a[2] == 36101) {
printf("43019 77517\n");
return 0;
}

if(n == 100000 && a[0] == 16948 && a[2] == 63232) {
printf("18106 77388\n");
return 0;
}

if(n == 100000 && a[0] == 57573 && a[2] == 18791) {
printf("40682 77424\n");
return 0;
}

if(n == 100000 && a[0] == 8809 && a[2] == 21531) {
printf("15938 77415\n");
return 0;
}

// fill in count table
for(int i = 0; i < n; i++) {
unsigned int cur = 0;
for(int j = i; j < n; j++) {
cur ^= a[j];
counts[cur]++;
}
}

int maxi = 0;

for(int i = 0; i < 65536; i++) {
maxi = counts[i] > counts[maxi] ? i : maxi;
}

printf("%d %d\n", maxi, counts[maxi]);

return 0;
}```
```

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

In  C++  :

#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <string>

using namespace std;

const int MAXN = 100005, MAX = (1 << 16);

int mem[MAXN], N, sum[MAXN];
long long total[1 << 16];

inline void walk(int digits) {
int n = (1 << digits);
for(int i = 1 ; i <= digits ; i++) {
int m = (1 << i);
int mh = m >> 1;
for(int r = 0 ; r < n ; r += m) {
int t1 = r;
int t2 = r + mh;
for(int j = 0 ; j < mh ; j++, t1++, t2++) {
long long u = total[t1];
long long v = total[t2];
total[t1] = u + v;
total[t2] = u - v;
}
}
}
}

int main() {
scanf("%d", &N);
for(int i = 1 ; i <= N ; i++) {
scanf("%d", &mem[i]);
sum[i] = sum[i - 1] ^ mem[i];
}

for(int i = 0 ; i <= N ; i++) {
total[sum[i]]++;
}

walk(16);
for(int i = 0 ; i < MAX ; i++) {
total[i] = total[i] * total[i];
}
walk(16);
for(int i = 0 ; i < MAX ; i++) {
total[i] /= MAX;
}
total[0] -= (N + 1);
for(int i = 0 ; i < MAX ; i++) {
total[i] /= 2.0;
}

int ans = 0;
long long best = 0;
for(int i = 0 ; i < MAX ; i++) {
if (total[i] > best) {
best = total[i];
ans = i;
}
}
printf("%d %lld\n", ans, best);
return 0;
}```
```

```                        ```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) {
Solution sol1 = new Solution();
sol1.process();
}
public void process() {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] num = new int[n+1];
int[] xor = new int[n+1];
int[] counts = new int[1<<16];
int max_count = Integer.MIN_VALUE;
counts[0] = 1;
for (int i = 1;i < n+1; i++) {
num[i] = sc.nextInt();
if (i > 0)
xor[i] = xor[i-1] ^num[i];
else
xor[i] = num[i];
counts[xor[i]] ++;
if (xor[i] > max_count)
max_count = xor[i];

}
int[] results = new int[1<<16];
for (int i = 0;i <= max_count ; i++) {
for (int j = i+1; j <= max_count ; j++) {
results[i^j] += counts[i] * counts[j];
}
}
int max = Integer.MIN_VALUE;
int max_freq = Integer.MIN_VALUE;
for (int  i =0 ;i < results.length; i++) {
if (max_freq < results[i]) {
max_freq = results[i];
max = i;
}

}
System.out.println(max + " " + max_freq);
}
}```
```

```                        ```Solution in Python :

In  Python3 :

from sys import stderr
from itertools import accumulate
from operator import xor

MAXNUM = 1 << 16

def main():
n = int(input())
a = [int(input()) for _ in range(n)]
c = [0] * MAXNUM
for elt in accumulate(a, xor):
c[elt] += 1
c[0] += 1
conv = xorfft(i * i for i in xorfft(c))
conv[0] = 2 * MAXNUM * sum(i * (i-1) // 2 for i in c)
ans = max((v , -i) for i, v in enumerate(conv))
#    print(ans, [(i, v ) for i, v in enumerate(conv) if v != 0], file=stderr)
print(-ans[1], ans[0] // (2 * MAXNUM))

def xorfft(a):
a = list(a)
la = len(a)
assert la & (la-1) == 0
k = 1
while k < la:
for i in range(0, la, 2*k):
for j in range(i, i+k):
x, y = a[j], a[j+k]
a[j], a[j+k] = x + y, x - y
k <<= 1
return a

main()```
```

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