题目：
Given an array nums of distinct positive integers, return the number of tuples (a, b, c, d) such that a * b = c * d where a, b, c, and d are elements of nums, and a != b != c != d.
Example 1:
Input: nums = [2,3,4,6] Output: 8 Explanation: There are 8 valid tuples: (2,6,3,4) , (2,6,4,3) , (6,2,3,4) , (6,2,4,3) (3,4,2,6) , (4,3,2,6) , (3,4,6,2) , (4,3,6,2)
Example 2:
Input: nums = [1,2,4,5,10] Output: 16 Explanation: There are 16 valids tuples: (1,10,2,5) , (1,10,5,2) , (10,1,2,5) , (10,1,5,2) (2,5,1,10) , (2,5,10,1) , (5,2,1,10) , (5,2,10,1) (2,10,4,5) , (2,10,5,4) , (10,2,4,5) , (10,2,4,5) (4,5,2,10) , (4,5,10,2) , (5,4,2,10) , (5,4,10,2)
Example 3:
Input: nums = [2,3,4,6,8,12] Output: 40
Example 4:
Input: nums = [2,3,5,7] Output: 0
Constraints:
1 <= nums.length <= 10001 <= nums[i] <= 10^4- All elements in
numsare distinct.
思路：
首先需要观察出对于给定4个数，要使abcd拜访成立，有且只有8种方法，这个从例子种就可以看出。因为数量级为10的4次，显然能够接受二重循环。接下来基本和找出相同积的对数和相同了，对于第i个数字，遍历0到i-1，如果这两个数字的积在哈希表中，则我们记录答案，这里需要将哈希表中相同积出现次数乘8，然后将当前积更新如哈希表中即可。
代码：
class Solution {
public:
int tupleSameProduct(vector<int>& nums) {
unordered_map<int,int> m;
int count=0;
for(int i=0;i<nums.size();i++)
{
for(int j=0;j<i;j++)
{
int temp=nums[i]*nums[j];
if(m.count(temp))
count+=8*m[temp];
m[temp]++;
}
}
return count;
}
};