# Flipping bits

### Problem Statement :

```You will be given a list of 32 bit unsigned integers. Flip all the bits ( 1 -> 0 and 0 -> 1 ) and return the result as an unsigned integer.

Function Description

Complete the flippingBits function in the editor below.

flippingBits has the following parameter(s):

int n: an integer
Returns

int: the unsigned decimal integer result
Input Format

The first line of the input contains q, the number of queries.
Each of the next  q lines contain an integer, n, to process.

Constraints

1  <=   q   <=  100
1   <=   n  <=  2^32

Sample Input 0

3
2147483647
1
0

Sample Output 0

2147483648
4294967294
4294967295```

### Solution :

```                            ```Solution in C :

In   C :

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

int main() {

int t;
unsigned int n;
scanf("%d", &t);
while(t-- > 0) {
scanf("%u", &n);
printf("%u\n", ~n);
}
return 0;
}```
```

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

In    C++ :

#include <cstdio>
using namespace std;

int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i ++) {
unsigned int val;
scanf("%u", &val);
printf("%u\n", ~val);
}
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) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
long mask = 0x00000000ffffffff;

for(int i = 0; i < n; i++) {
long l = sc.nextLong();
}
}
}```
```

```                        ```Solution in Python :

In   Python3  :

T = int(input())
for _ in range(T) :
N = int(input())
print(0xffffffff & ~N)```
```

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