# 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.

Example
n=9(10)
9(10)=1001(2). We're working with 32 bits, so:
000000000000000000000000000001001 = 9(10)
111111111111111111111111111110110 = 4294967286(10)
Return 4294967286.

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
0 <= n < 2^32```

### Solution :

```                            ```Solution in C :

In C++ :

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

ll val = 4294967295L;
int main() {
int T;
cin >> T;
for(int t=0;t<T;t++){
ll cur;
cin >> cur;
cout << (val-cur) << endl;
}
return 0;
}

In Java :

import java.util.Scanner;

public class Solution {

public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);

int T = scanner.nextInt();

long mask = (1L << 32) - 1;

for (int t = 0; t < T; t++) {
long n = scanner.nextLong();
}

scanner.close();

}

}

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;
}

In Python3 :

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

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