# PCJ18A - Editorial

Tester: Prakhar Gupta
Editorialist: Prakhar Gupta

CAKEWALK

None

### PROBLEM:

Given N dishes, find out if there is atleast one dish with amount of secret ingredient \geq X

### EXPLANATION:

We go over the amount of secret ingredient in all the N dishes. If we find a dish which has amount of secret ingredient more than or equal to X, we print ‘YES’.

One mistake participants did was break from the input loop once they found an A_i \geq X. This led to the remaining input of the current test case to be transferred to the next test case, leading to a Wrong Answer, or sometimes a Runtime Error verdict.

Complexity: The time complexity is O(N) per test case.

### AUTHOR’S SOLUTION:

Author’s solution can be found here.

2 Likes
``````#include <iostream>
using namespace std;
using ll = long long;

int main() {
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
int ans[1001];
for (int i = 0; i < n; i++) {
cin >> ans[i];
}
bool an = false;
for (int i = 0; i < n; i++) {
if (ans[i] >= x) {
an = true;
}
}
if (an) {
cout << "YES" << endl;
}
else {
cout << "NO" << endl;
}
}
}
``````