CF 105314J - Ahmad and Prediction Syndrome

CF 105314J - Ahmad and Prediction Syndrome

Rating: -
Tags: -
Solve time: 39s
Verified: yes

Solution

Problem Understanding

The task describes a very simple qualification rule for a contest. Each test case gives a single integer representing how many distinct balloons a team has robbed. A team qualifies if and only if it has robbed at least 8 different balloons.

So for every input number, we are not simulating anything or building a structure. We are only checking whether that number reaches a fixed threshold. The output is a sequence of answers, one per test case, indicating whether the condition is satisfied.

The constraints are extremely small. The number of test cases is at most 10, and each value of n is at most 10. This immediately rules out any concern about performance, memory usage, or optimization techniques. Even a solution that performs unnecessary work would still pass comfortably because the total input size is constant in practice.

There are no subtle edge cases in the classical sense, but there is one implementation pitfall that appears often in problems like this: mixing strict and non-strict comparisons. A naive attempt might check n > 8 instead of n >= 8, which would incorrectly reject the boundary case where n equals 8. For example, input 8 should output YES, but a strict inequality would incorrectly output NO.

Another possible mistake is output formatting. Since each test case must be printed on its own line, concatenating results without line breaks would produce a wrong answer even though the logic is correct.

Approaches

A brute-force interpretation would try to model the problem more generally: one might imagine iterating over some representation of balloons or simulating the act of “robbing distinct items.” In a realistic generalization, this could involve maintaining a set of items and checking its size. That approach would look like inserting elements into a set and checking whether its cardinality reaches 8. In this particular problem, however, that entire process is already completed for us in the input.

The key observation is that the input integer already encodes the final set size. There is no hidden structure to reconstruct, no duplicates to filter, and no sequence to process. Each test case is independent and fully self-contained.

So the solution collapses into a single comparison per test case. We compare n against the threshold 8 and print the corresponding answer. The brute-force idea of building a set is conceptually correct but unnecessary overhead for a value that is already given directly.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (simulate set)	O(t)	O(1)	Accepted (overkill)
Optimal (direct check)	O(t)	O(1)	Accepted

Algorithm Walkthrough

1. Read the number of test cases t. This determines how many independent checks we will perform.
2. For each test case, read the integer n representing how many distinct balloons were robbed.
3. Compare n with 8.
4. If n is greater than or equal to 8, output "YES". Otherwise, output "NO".

Each step mirrors the structure of the problem directly. There is no preprocessing or state carried between test cases, so each iteration is fully independent.

Why it works

The problem definition reduces qualification to a single threshold condition on a scalar value. Since n already represents the count of distinct balloons, it fully captures all relevant information. The decision boundary at 8 partitions the entire input space into two disjoint sets: all values below 8 are invalid, and all values from 8 upward are valid. Because the comparison is exhaustive and mutually exclusive, every input maps to exactly one correct output.

Python Solution

import sys
input = sys.stdin.readline

t = int(input().strip())
for _ in range(t):
    n = int(input().strip())
    if n >= 8:
        print("YES")
    else:
        print("NO")

The solution reads input efficiently using sys.stdin.readline, although efficiency is not critical here given the tiny constraints. Each test case is processed in constant time.

The only important implementation detail is the boundary condition in the comparison. Using >= 8 is essential because the threshold is inclusive. Any strict comparison would misclassify the boundary value.

Worked Examples

Example 1

Input:

4
1
3
7
8

Test	n	Condition (n ≥ 8)	Output
1	1	False	NO
2	3	False	NO
3	7	False	NO
4	8	True	YES

This trace shows how the decision flips exactly at the boundary value 8. Everything below remains in the rejecting region, while 8 and above qualify.

Example 2

Input:

3
10
8
9

Test	n	Condition (n ≥ 8)	Output
1	10	True	YES
2	8	True	YES
3	9	True	YES

This case confirms that all values in the valid range behave identically. Once the threshold is crossed, no further distinctions matter.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(t)	Each test case requires one constant-time comparison
Space	O(1)	No auxiliary structures are used

The constraints cap t at 10 and n at 10, so this solution runs instantly and uses negligible memory, far below any practical limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    input = sys.stdin.readline

    t = int(input().strip())
    out = []
    for _ in range(t):
        n = int(input().strip())
        if n >= 8:
            out.append("YES")
        else:
            out.append("NO")
    return "\n".join(out)

# provided sample (reconstructed behavior)
assert run("4\n1\n3\n7\n8\n") == "NO\nNO\nNO\nYES"

# minimum value
assert run("1\n1\n") == "NO"

# boundary check
assert run("1\n8\n") == "YES"

# above boundary
assert run("1\n10\n") == "YES"

# mixed case
assert run("5\n0\n8\n9\n7\n2\n") == "NO\nYES\nYES\nNO\nNO"

Test input	Expected output	What it validates
1, 1	NO	Minimum edge case
8	YES	Boundary correctness
10	YES	Above-threshold correctness
mixed values	mixed	Consistency across cases

Edge Cases

The only meaningful edge case is the threshold boundary at n = 8. For input:

1
8

The algorithm reads t = 1, then n = 8. It evaluates the condition 8 >= 8, which is true, so it outputs YES. This confirms that the boundary is included correctly.

A strict inequality implementation would instead evaluate 8 > 8 as false and output NO, which would violate the problem requirement.