Sum of Floored Pairs
<p>Given an integer array <code>nums</code>, return the sum of <code>floor(nums[i] / nums[j])</code> for all pairs of indices <code>0 &lt;= i, j &lt; nums.length</code> in the array. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>The <code>floor()</code> function returns the integer part of the division.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,5,9]
<strong>Output:</strong> 10
<strong>Explanation:</strong>
floor(2 / 5) = floor(2 / 9) = floor(5 / 9) = 0
floor(2 / 2) = floor(5 / 5) = floor(9 / 9) = 1
floor(5 / 2) = 2
floor(9 / 2) = 4
floor(9 / 5) = 1
We calculate the floor of the division for every pair of indices in the array then sum them up.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [7,7,7,7,7,7,7]
<strong>Output:</strong> 49
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>
Array; Math; Binary Search; Prefix Sum
func sumOfFlooredPairs(nums []int) (ans int) {
	mx := slices.Max(nums)
	cnt := make([]int, mx+1)
	s := make([]int, mx+1)
	for _, x := range nums {
		cnt[x]++
	}
	for i := 1; i <= mx; i++ {
		s[i] = s[i-1] + cnt[i]
	}
	const mod int = 1e9 + 7
	for y := 1; y <= mx; y++ {
		if cnt[y] > 0 {
			for d := 1; d*y <= mx; d++ {
				ans += d * cnt[y] * (s[min((d+1)*y-1, mx)] - s[d*y-1])
				ans %= mod
			}
		}
	}
	return
}