Total Hamming Distance

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Now your job is to find the total Hamming distance between all pairs of the given numbers.

Example:

`Input: 4, 14, 2 Output: 6 Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just showing the four bits relevant in this case). So the answer will be: HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6. `

Note:

1. Elements of the given array are in the range of `0 `to `10^9`
2. Length of the array will not exceed `10^4`.
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Using bit manipulation to achieve operators

1）计算负数的逆元

```def negative(n):
return bit_add(~n, 1)```

2）加法计算

```def bit_add(a, b):
b &= (2 ** 32 - 1)
while b:
tmp = a
sum = tmp ^ b
carry = (tmp & b) << 1
return a```

3）减法计算

```def bit_sub(a, b):
return bit_add(a, negative(b))```

4）乘法计算

```def bit_mul(a, b):
result = 0

while b:
if b & 0x1:
result = bit_add(result, a)
b >>= 1
a <<= 1

return result```

5）除法计算

```def bit_div(x, y):
result = 0
for i in range(32)[::-1]:
if (x >> i) >= y:
result += (1 << i)
x -= (y << i)

return result```