Given an array of integers A, find the number of triples of indices (i, j, k) such that:

• 0 <= i < A.length
• 0 <= j < A.length
• 0 <= k < A.length
• A[i] & A[j] & A[k] == 0, where & represents the bitwise-AND operator.

Example 1:

Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
 
 

Note:

1. 1 <= A.length <= 1000
2. 0 <= A[i] < 2^16

給定一個整數數組 A，找出索引為 (i, j, k) 的三元組，使得：

• 0 <= i < A.length
• 0 <= j < A.length
• 0 <= k < A.length
• A[i] & A[j] & A[k] == 0，其中 & 表示按位與（AND）操作符。

示例：

輸入：[2,1,3]
輸出：12
解釋：我們可以選出如下 i, j, k 三元組：
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2 

提示：

1. 1 <= A.length <= 1000
2. 0 <= A[i] < 2048

9888ms

 1 class Solution {
 2     func countTriplets(_ A: [Int]) -> Int {
 3         var cnt:[Int] = [Int](repeating:0,count:65536)
 4         for i in 0..<65536
 5         {
 6             for n in A
 7             {
 8                 if
 
 (i&n)==0
 9                 {
10                     cnt[i] += 1
11                 }
12             }
13         }
14         
15         var res:Int = 0
16         for a in A
17         {
18             for b in A
19             {
20                 res += cnt[a&b]
21             }
22         }
23         return res
24     }
25 }

[Swift Weekly Contest 121]LeetCode982. 按位與為零的三元組 | Triples with Bitwise AND Equal To Zero