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

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

本来是个很简单的题目，不小心我给写复杂了。

但这样就很好理解了。我们找的就是余数也是i和60-i这种，然后考虑排列组合就好了。注意0和30这种组合。

简单的代码依然可以看大佬的

class Solution {

public:

    int numPairsDivisibleBy60(vector<int>& time) {

        int count[] = {};

        int len = time.size();

        for(int i =  ; i < len ; i++){

            int x = time[i] % ;

            count[x]++;

        }

        int i;

        int result = count[] * (count[] - ) / ;

        for(i =  ;i < ( + ) / ; ++ i){

            result += count[i] * count[ - i];

        }

        if( * i == ){

            result += count[i] * (count[i] - ) / ;

        }

        return result;

    }

};