<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 n=nums.size();
        vector<pair<int,int>> dp(n,{0,0});
        vector<int> s(n,0);
        for(int i=0; i<n; i++){
            int c=0, d=1;
            for(int j=0; j<i; j++){
                if(nums[j]<nums[i]){
                    if(dp[j].first>c){
                        c=dp[j].first; d=dp[j].second;
                    }else if(dp[j].first==c){
                        d+=dp[j].second;
                    }
                }
            }
            dp[i].first=c+1;
            dp[i].second=d;
        }
        for(auto i:dp) cout<<i.first<<"-"<<i.second<<" ";
        cout<<endl;
        sort(dp.begin(),dp.end());
        int t=dp[n-1].second;
        for(int i=n-2; i>=0; i--){
            if(dp[i].first==dp[i+1].first) t+=dp[i].second;
            else break;
        }
        return t;
    }
};
16600