Pairs of SongsWithTotalDurationsDivisibleby60*

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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
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class Solution {
public:
	int numPairsDivisibleBy60(vector<int>& time) {
		int flag[60], ret = 0, tmp;
		for (int i = 0; i<60; ++i)flag[i] = 0;
		for (auto t : time) {
			tmp = t % 60;
			ret += flag[(60 - tmp) % 60];
			flag[tmp] += 1;
		}
		return ret;
	}
};