Switchblade of Thanos

Notice that the maximum sum of all elements can be at most $10^5$. So, we can easily use Dynamic Programming to solve this problem.

Let $dp1[i][j]$ be the number of subsequences with sum $j$ ended at or before the $i$-th element. We either take the $i$-th element for this subsequence of not take it. So the transition is $dp1[i][j] = dp1[i-1][j-A[i]] + dp1[i-1][j]$.

Again, let $dp2[i][j]$ be the number of subsequences with sum $j$started from or after the $i$-th element.

The transition for this case is $dp2[i][j] = dp2[i+1][j-A[i]] + dp2[i+1][j]$.

So the number of subsequences with sum $2*j$ with the second equal part starting at the $i$-th element is $dp1[i-1][j] * dp2[i+1][j-A[i]]$.

Solve this for all $i$ and $j$ and the summation of the results is the desired answer.