In this problem, you just need a simple math formula that is just given inside the problem statement. Thus the formula: $$$a*3 + b * 1 + c * 0$$$

#include <bits/stdc++.h>
using namespace std;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL);
#define endl "\n"

int main() {
    fastio;
    int a, b, c; cin >> a >> b >> c;
    cout << (a * 3 + b) << endl; // because C cancels out and b * 1 is b.
}

In this problem, we need to keep track of the adjacent elements in the array and find the smallest in them. And now, we want the max possible height so, the answer is $$$maximum = max(maximum, min(a[i], a[i + 1]))$$$

#include <bits/stdc++.h>
using namespace std;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL);
#define endl "\n"

int main() {
    fastio;
    int t; cin >> t;
    while(t--){
        int n; cin >> n; 
        vector<int> a(n);
        for(int i = 0; i<n;i++) cin >> a[i];
        int maxh = 0;
        for(int i = 0; i < n-1; i++){
            int curr = min(a[i], a[i + 1]); // keeping track of adjacent elements
            maxh = max(curr, maxh);
        }
        cout << maxh << endl;
    }
}

So, In this problem, each red card results in the footballer begin sent off so at min $$$R$$$ send offs. We have to minimize this no. of send offs. Each red card can free 1 yellow card up. So we can cancel out upto R yellow cards in the order $$$(yellow, red, yellow, red, yellow, red, …)$$$

So assume Y = yellow, R = red. if Y is lesser than equal to R. in this case, we'll be able to cancel out all of the yellow cards. So, min total no. of send-offs will equal exactly R, R — one from each red card.

The outcome of this will result in the formula:

• $$$R + \dfrac{\max(0, y - r)}{2}$$$ for any testcase.

#include <bits/stdc++.h>
using namespace std;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL);
#define endl "\n"

int main() {
    fastio;
    int t; cin >> t;
    while(t--){
        int r, y; cin >> r >> y;
        cout << r + (max(0, y - r) / 2) << endl;
    }
}