题目如下：

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 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. 1 <= time.length <= 60000
2. 1 <= time[i] <= 500

解题思路：遍历Input并对其中每个元素与60取模，以余数为key值存入字典dic中，字典的value也key值作为余数出现的次数。接下来再遍历一次Input，求出元素与60取模后的余数，再求出60减去余数的差值，字典dic[差值]所对应的值即为这个元素可以与数组中多少个元素的和能被60整除。

代码如下：

class Solution(object):

    def numPairsDivisibleBy60(self, time):

        """

        :type time: List[int]

        :rtype: int

        """

        dic = {}

        for i in time:

            v = i % 60

            dic[v] = dic.setdefault(v,0) + 1

        res = 0

        for i in time:

            v = i % 60

            dic[v] -= 1

            key = 60 -v if v != 0 else 0

            if key in dic:

                res += dic[key]

        return res