<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>

class Solution {
public:
    int findNumberOfLIS(vector<int>& nums) {
        int cnt = 0, n = nums.size(), mlen = 1;
        vector<int> dp(n, 0), cp(n, 0);
        dp[0] = cp[0] = 1;
        for (int i = 1; i < n; i++){
            dp[i] = cp[i] = 1;
            for (int j = 0; j < i; j++){
                if (nums[j] < nums[i]){
                    if (dp[i] == 1 + dp[j]){
                        cp[i] += cp[j];
                    } else if (dp[i] <= dp[j]){
                        dp[i] = 1 + dp[j];
                        cp[i] = cp[j];
                    }
                }
            }
            mlen = max(mlen, dp[i]);
        }
        for (int i = 0; i < n; i++){
            if (dp[i] == mlen) cnt += cp[i];
        }
        return cnt;
    }
};
16300