Since we got the first dice and the result of it, the second dice should be in the range of 6 if subtracted from 9 since, we love the number 9 and both the numbers add up to it.

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

int main() {
    int x; cin >> x;
    
    int n = 9 - x;
    cout << (n <= 6 ? "YES" : "NO") << endl; 
}

This problem is pretty straight-forward so, just catch the frequency of every $$$n \gt = 60$$$ and $$$n \gt = 30$$$ and just check if the frequency of the 60s is $$$ \gt =2$$$ and 30s is $$$ \gt =4$$$

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

int main() {
    int t; cin >> t;
    while(t--){
        int n = 5;
        vector<int> a(n);
        for(int i = 0; i < n; i++) cin >> a[i];
        
        int sixetys = 0, thirtys = 0;
        for(int i = 0; i < n; i++){
            if(a[i] >= 60) sixetys++;
            if(a[i] >= 30) thirtys++;
        }
        cout << (sixetys >= 2 && thirtys >= 4 ? "Pass" : "Fail") << endl; 
    }
}

This problem hides a simple trick to just identify the $$$max(arr)$$$ and check that how many elements are equal to it.

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

int main() {
    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 maxs = *max_element(a.begin(), a.end());
        
        int l = 0;
        for(int i = 0; i < n; i++){if(a[i]==maxs) l++;}
        cout << l << endl;
    }
}