human
I found an interesting problem on https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/:
In a list of songs, the i-th song has a duration of time[i] seconds. 
Return the number of pairs of songs for which their total duration in seconds is divisible by 60.  Formally, we want the number of indices i, j such that i < j with (time[i] + time[j]) % 60 == 0.
 
Example 1:
Input: [30,20,150,100,40]
Output: 3
Explanation: Three pairs have a total duration divisible by 60:
(time[0] = 30, time[2] = 150): total duration 180
(time[1] = 20, time[3] = 100): total duration 120
(time[1] = 20, time[4] = 40): total duration 60

Example 2:
Input: [60,60,60]
Output: 3
Explanation: All three pairs have a total duration of 120, which is divisible by 60.

 
Note:

1 <= time.length <= 60000
1 <= time[i] <= 500

I tried it in Python, but could not do it. Can you solve it?

 ... Can you provide a the Python program you requested?
assistant
#Here is the Python program you requested.

from collections import Counter
class Solution:
    def numPairsDivisibleBy60(self, time: List[int]) -> int:
        count = Counter(time)
        pair = 0
        for p in count :
            tp = count[p]
            t = 60 - p % 60
            while t <= 500 :
                pair += tp*count[t] if t != p else tp * (tp-1)
                t += 60
        
        return pair // 2