【leetcode】1013. Pairs of Songs With Total Durations Divisible by 60

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