1010. Pairs of Songs With Total Durations Divisible by 60

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本文介绍了一种高效算法，用于解决歌曲配对问题，即找出歌曲列表中总时长能被60整除的歌曲对。通过巧妙利用补集思想，算法仅需遍历一次列表，大大提高了效率。

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

难度：easy
解题思路：本题用暴力法时间复杂度是O(n^2)，会超时，本题借用补集的思想，遍历一次即可。
解题代码：

class Solution {
public:
    int a[510];
    int numPairsDivisibleBy60(vector<int>& time) {
        
       int ret=0;
        for(int i=0;i<time.size();i++){
            ret+=a[(60-time[i]%60)%60];
            a[time[i]%60]++;
        }
        return ret;
    }
};