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You are given a list of songs where the `ith` 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` such that `i < j` with `(time[i] + time[j]) % 60 == 0`.

Example 1:

```Input: time = [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: time = [60,60,60]
Output: 3
Explanation: All three pairs have a total duration of 120, which is divisible by 60.
```

Constraints:

• `1 <= time.length <= 6 * 104`
• `1 <= time[i] <= 500`

`Code : `

``````    int numPairsDivisibleBy60(vector<int>& time) {
vector<int> c(60);
int res = 0;
for (int t : time) {
res += c[(600 - t) % 60];
c[t % 60] += 1;
}
return res;
}``````
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