HackerRank XORing Ninja problem solution

In this HackerRank XORing Ninja problem solution, we have given a list of N space-separated integers and we need to determine and print the XORSUM value. and the XORSUM is equal to the summation of the XOR of the array.

Problem solution in Python.

M=1000000007
tests = int(input().strip())
for i in range(0,tests):
    n = int(input().strip())
    a = [int(x.strip()) for x in input().strip().split()]
    b0 = [0 for y in range(0,32)]
    b1 = [0 for y in range(0,32)]
    for k in range(0,n):        
        for j in range(0,32):
            if(a[k] & (1<<j)):
                tmp = b1[j]
                b1[j]=(b1[j]+1+b0[j])%M
                b0[j]=(b0[j]+tmp)%M
            else:
                b1[j]=(b1[j]+b1[j])%M
                b0[j]=(1+b0[j]+b0[j])%M
        

    cum = 0
    for j in range(0,32):
        val = ((1<<j)*b1[j])%M
        cum=(cum+val)%M

    print(cum)

Problem solution in Java.

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

public class Solution {

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner s = new Scanner(System.in);
        int numTests = s.nextInt();
        for(int tests = 0; tests < numTests; tests++){
            int n = s.nextInt();
            long sum = 0;
            for(int i = 0; i < n; i++){
                sum = sum|s.nextLong();
            }
            for(int i = 0; i < n-1; i++){
                sum = sum*2 % (1000000007);
            }
            long finalSum = sum << (n-1);
            long finalMod = finalSum % (1000000007);
            System.out.println(sum);
        }
    }
}

Problem solution in C++.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


const int M = 1000000007;

int p(int x) {
	if (x == 0) {
       return 1;
    }	
  	long long y = p(x >> 1);
  	int r = y * y % M;
  	if (x & 1) {
     	r <<= 1;
    }
  	return (r >= M)?(r - M):r; 
}

int main() {
 int z;
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */  
  	for (scanf("%d",&z);z;--z) {
      int state = 0,n;
      scanf("%d",&n);
      
    	for (int i = 0; i < n; ++i) {
        	int x = 0;
          	scanf("%d",&x);
        	for (int j = 0; (1 << j) <= x; ++j) {
            	if (x & (1 << j)) {
                  	state |= (1 << j);
                }
              
            }  
        }
      	int r = 0;
        for (int j = 0; j <= 30; ++j) {
          	if ((state & (1 << j)) && ((r += p(n - 1 + j)) >= M)) {
              	r -= M;
            }
          
        }
      	printf("%d\n",r);
    }
    return 0;
}

Problem solution in C.

#include<stdio.h>
#define LL long long int
#define MOD 1000000007

LL power(LL a, LL b)
{
	if (b == 0)
		return 1;
	else
	{
		LL temp=(power(a,b/2))%MOD;
		if(b%2==0)   
			return (temp*temp)%MOD;
		else
			return (((temp*a)%MOD)*temp)%MOD;
	}
}

int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		int n;
		scanf("%d",&n);
		int x=0;
		int i,a;
		for(i=0;i<n;i++)
		{
			scanf("%d",&a);
			x|=a;
		}
		LL ans=power(2,n-1);
		ans=(ans*x)%MOD;
		printf("%lld\n",ans);
	}
	return 0;
}