**Problem Idea**↵
↵
We areThe problem basically givens an array of positive integers (each value is between 1 and 67). For every test case, we need to decide if it’s possible to choose at least one element (a non-empty subset) such that the product of the chosen numbers becomes exactly 67.↵
If it’s possible, print “YES”; otherwise, print “NO”.↵
↵
**Key Observation**↵
↵
The whole problem becomes very easy once you notice one thing:↵
67 is a prime number.↵
A prime number has only two factors: 1 and itself.↵
So the only way to make the product 67 is:↵
Take the number 67 alone, or↵
Take 67 with some 1’s (because multiplying by 1 doesn’t change the product)↵
↵
That means:↵
If the array contains 67, we can just pick that single element, and the product will be 67.↵
If the array does not contain 67, then no combination of other numbers can multiply to 67 (because 67 can’t be formed by multiplying smaller numbers except using 1 and 67).↵
↵
**Final Approach**↵
↵
For each test case:↵
Read n and the array↵
Check if 67 exists in the array↵
If yes → print “YES”↵
Otherwise → print “NO”↵
↵
**Why This Works (Short Proof)**↵
↵
If 67 is present: choose {67} → product is 67 → valid↵
If 67 is not present: since 67 is prime, you cannot create it by multiplying other numbers → impossible↵
↵
**Complexity**↵
↵
The check is very small and fast: we only scan the array once.↵
So the time complexity is O(n) per test case (and here n is tiny anyway).↵
↵
That’s it, no sieve, no heavy math, just a simple prime observation.↵
↵
**Author:**↵
[user:ahmedmshakil,2026-02-21]
↵
If it’s possible, print “YES”; otherwise, print “NO”.↵
↵
**Key Observation**↵
↵
The whole problem becomes very easy once you notice one thing:↵
67 is a prime number.↵
A prime number has only two factors: 1 and itself.↵
So the only way to make the product 67 is:↵
Take the number 67 alone, or↵
Take 67 with some 1’s (because multiplying by 1 doesn’t change the product)↵
↵
That means:↵
If the array contains 67, we can just pick that single element, and the product will be 67.↵
If the array does not contain 67, then no combination of other numbers can multiply to 67 (because 67 can’t be formed by multiplying smaller numbers except using 1 and 67).↵
↵
**Final Approach**↵
↵
For each test case:↵
Read n and the array↵
Check if 67 exists in the array↵
If yes → print “YES”↵
Otherwise → print “NO”↵
↵
**Why This Works (Short Proof)**↵
↵
If 67 is present: choose {67} → product is 67 → valid↵
If 67 is not present: since 67 is prime, you cannot create it by multiplying other numbers → impossible↵
↵
**Complexity**↵
↵
The check is very small and fast: we only scan the array once.↵
So the time complexity is O(n) per test case (and here n is tiny anyway).↵
↵
That’s it, no sieve, no heavy math, just a simple prime observation.↵
↵
**Author:**↵
[user:ahmedmshakil,2026-02-21]
