# 1362. Closest Divisors

Given an integer `num`, find the closest two integers in absolute difference whose product equals `num + 1` or `num + 2`.

Return the two integers in any order.

**Example 1:**

```
Input: num = 8
Output: [3,3]
Explanation: For num + 1 = 9, the closest divisors are 3 & 3, for num + 2 = 10, the closest divisors are 2 & 5, hence 3 & 3 is chosen.
```

**Example 2:**

```
Input: num = 123
Output: [5,25]
```

**Example 3:**

```
Input: num = 999
Output: [40,25]
```

**Constraints:**

* `1 <= num <= 10^9`

对给定的整数num，找两个最近的整数使得它们的积等于num + 1或num + 2。当num + 1或2有平方根时，最近的两个数是平方根。因此从平方根开始往下遍历，遇到的第一个能被整除的就是结果。又因为num+1比+2小，当除数为1时返回num + 1能得到差的绝对值更小的。

```cpp
class Solution {
public:
    vector<int> closestDivisors(int num) {
        for (int i = sqrt(num + 2); i >= 1; --i) {
            if ((num + 1) % i == 0) return {(num + 1) / i, i};
            if ((num + 2) % i == 0) return {(num + 2) / i, i};
        };
        return {};
    }
};
```


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