#Leetcode# 1010. Pairs of Songs With Total Durations Divisible by 60

https://leetcode.com/problems/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

class Solution {
public:
    int numPairsDivisibleBy60(vector<int>& time) {
        int num[100];
        memset(num, 0, sizeof(num));
        int len = time.size();
        int ans = 0;
        for(int i = 0; i < len; i ++) 
            num[time[i] % 60] ++;
        for(int i = 1; i < 30; i ++) 
            ans += num[i] * num[60 - i];
        if(num[30]) {
            ans += num[30] * (num[30] - 1) / 2;
        }
        if(num[0]) {
            ans += num[0] * (num[0] - 1) / 2;
        }
        return ans;
    }
};