**Sum vs XOR**

### Problem Statement :

Given an integer , find each such that: 0 <= x <= n n + x = n+x where + denotes the bitwise XOR operator. Return the number of x's satisfying the criteria. Example n = 4 There are four values that meet the criteria: 4 + 0 = 4 + 0 = 4 4 + 1 = 4 + 1 = 5 4 + 2 = 4 + 2 = 6 4 + 3 = 4 + 3 = 7 Return 4. Function Description Complete the sumXor function in the editor below. sumXor has the following parameter(s): - int n: an integer Returns - int: the number of values found Input Format A single integer, n. Constraints 0 <= n <= 10^15

### Solution :

` ````
Solution in C :
In C++ :
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
int main(){
long long n;
scanf("%lld", &n);
long long res=1;
while(n){
if(n%2 == 0) res *= 2;
n /= 2;
}
printf("%lld", res);
return 0;
}
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 in = new Scanner(System.in);
long n = in.nextLong();
int zeroCount = 0;
while (n>0) {
if (n%2==0)
zeroCount++;
n/=2;
}
System.out.println(1l<<zeroCount);
}
}
In C :
#include <stdio.h>
int main(){
long long int n,m=1;
scanf("%lld",&n);
while(n>0){
if(n%2==0)m*=2;
n/=2;
}
printf("%lld\n",m);
return 0;
}
In Python3 :
#!/bin/python3
import sys
n = int(input().strip())
print(1 if n == 0 else 1 << (bin(n)[2:].count('0')))
```

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