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

You are given a list of songs where 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`

such that `i < j`

with `(time[i] + time[j]) % 60 == 0`

.

**Example 1:**

Input:time = [30,20,150,100,40]Output:3Explanation: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:3Explanation:All three pairs have a total duration of 120, which is divisible by 60.

**Constraints:**

`1 <= time.length <= 6 * 10`

^{4}`1 <= time[i] <= 500`

## Rust Solution

```
struct Solution;
impl Solution {
fn num_pairs_divisible_by60(time: Vec<i32>) -> i32 {
let mut a: Vec<i32> = vec![0; 60];
let mut res = 0;
for x in time {
let count = a[((600 - x) % 60) as usize];
if count != 0 {
res += count;
}
a[(x % 60) as usize] += 1;
}
res
}
}
#[test]
fn test() {
let time = vec![30, 20, 150, 100, 40];
assert_eq!(Solution::num_pairs_divisible_by60(time), 3);
let time = vec![60, 60, 60];
assert_eq!(Solution::num_pairs_divisible_by60(time), 3);
}
```

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