**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();
System.out.println(n ^ mask);
}
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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