Let $$$dp(n)$$$ be an array/vector where $$$dp[i]$$$ denotes the minimum Blimps required for the $$$i^{th}$$$ city. $$$Σ dp[i]$$$ will give me the final answer. I will keep a counter to update $$$A[i]$$$ due to any prior Confetti Operations.

$$$Case 1.$$$ When $$$X≥Y$$$. Always Perform Operation 2.
$$$Case 2.$$$ When $$$X<Y$$$.
Let us first check if $$$X≥A[i]$$$. In this case, even if $$$X<Y$$$, its still better to perform Operation 2 as Number of Blimps required would be the same.
Else, we perform Operation 1 until we have a value less than $$$Y$$$. We would now have $$$dp[i] = A[i] / Y$$$ and $$$A[i]$$$ $$$=$$$ $$$A[i]$$$ $$$mod$$$ $$$Y$$$.
If $$$A[i] != 0$$$, we again check if $$$X≥A[i]$$$. If the condition now holds, we perform Operation 2, (1 time) in addition to previous Operation 1's. If not, then we perform one more Operation 1 to get $$$dp[i]$$$.

#include <iostream>
#include <vector>
using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int tc; cin >> tc;
    while(tc--)
    {
        int n, x, y;
        cin >> n >> x >> y;

        vector<int> v(n);
        for(int i=0; i<n; i++) cin >> v[i];

        vector<int> dp(n);
        int ctr = 0;

        for(int i=n-1; i>=0; i--)
        {
            v[i] -= ctr; // update unhappiness

            if(v[i] <= 0) dp[i] = 0;  // already happy

            else
            {
                if(x >= y) // always drop confetti
                {
                    dp[i] = v[i] / x;
                    dp[i] = ((v[i] % x) != 0) ? dp[i] + 1 : dp[i];
                    ctr += y * dp[i];
                }

                else
                {
                    if(x >= v[i]) // use confetti even if y > x
                    {
                        dp[i] = 1;
                        ctr += y;
                    } 
                    else
                    {
                        dp[i] = v[i] / y;
                        v[i] = v[i] % y; // remaining

                        if(v[i] != 0)
                        {
                            if(x >= v[i]) ctr += y; // use confetti even if y > x
                            dp[i] += 1;
                        }
                    }
                }
            }
        }

        int s = 0;
        for(int i=0; i<n; i++) s += dp[i];
        
        cout << s << '\n';
    }

    return 0;
}