128th LeetCode Weekly Contest 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 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

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[60] = {0};
        int len = time.size();
        for(int i = 0 ; i < len ; i++){
            int x = time[i] % 60;
            count[x]++;
        }
        int i;
        int result = count[0] * (count[0] - 1) / 2;
        for(i = 1 ;i < (60 + 1) / 2; ++ i){
            result += count[i] * count[60 - i];
        }
        if(2 * i == 60){
            result += count[i] * (count[i] - 1) / 2;
        }
        return result;
    }
};