Number of Longest Increasing Subsequence
<p>Given an integer array&nbsp;<code>nums</code>, return <em>the number of longest increasing subsequences.</em></p>

<p><strong>Notice</strong> that the sequence has to be <strong>strictly</strong> increasing.</p>

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

<pre>
<strong>Input:</strong> nums = [1,3,5,4,7]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7].
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,2,2,2,2]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The length of the longest increasing subsequence is 1, and there are 5 increasing subsequences of length 1, so output 5.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2000</code></li>
	<li><code>-10<sup>6</sup> &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
	<li>The answer is guaranteed to fit inside a 32-bit integer.</li>
</ul>
Binary Indexed Tree; Segment Tree; Array; Dynamic Programming
func findNumberOfLIS(nums []int) (ans int) {
	n, mx := len(nums), 0
	f := make([]int, n)
	cnt := make([]int, n)
	for i, x := range nums {
		for j, y := range nums[:i] {
			if y < x {
				if f[i] < f[j]+1 {
					f[i] = f[j] + 1
					cnt[i] = cnt[j]
				} else if f[i] == f[j]+1 {
					cnt[i] += cnt[j]
				}
			}
		}
		if mx < f[i] {
			mx = f[i]
			ans = cnt[i]
		} else if mx == f[i] {
			ans += cnt[i]
		}
	}
	return
}