004-002-002

Else If: Temperature Level Determination

Easy

Problem Description

Else If: Temperature Level Determination

Learning Objective: Perform staged classification with else if and evaluate multiple conditions in order

Determine comfort level from temperature. Display 5-level comfort message based on temperature.

Input

Line 1: Temperature (integer)

Output

Temperature: [temp] degrees
Level: [Extreme heat/Hot/Comfortable/Cold/Freezing]
```java

## Criteria
- 35 or higher: "Extreme heat"
- 25 to 34: "Hot"
- 15 to 24: "Comfortable"
- 5 to 14: "Cold"
- Below 5: "Freezing"

## Examples

### Example 1: Extreme Heat Level
Input:
```java
38
```java
Output:
```java
Temperature: 38 degrees
Level: Extreme heat
```java
38 >= 35 so "Extreme heat"

### Example 2: Comfortable Level
Input:
```java
18
```java
Output:
```java
Temperature: 18 degrees
Level: Comfortable
```java
15 <= 18 < 25 so "Comfortable"

### Example 3: Boundary Value (15)
Input:
```java
15
```java
Output:
```java
Temperature: 15 degrees
Level: Comfortable
```java
15 >= 15 so included in "Comfortable".

Test Cases

※ Output examples follow programming industry standards

Input:

Expected Output:

Temperature: 38 °C
Level: Extreme heat

Input:

Expected Output:

Temperature: 18 °C
Level: Comfortable

Input:

Expected Output:

Temperature: 15 °C
Level: Comfortable

❌ Some tests failed

❌ エラー発生

Your Solution

Current Mode:● My Code

import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    // Write your code here

    sc.close();
  }
}

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