** Solution to Problem 1807A – Plus or Minus**

Hello Codeforces!

Today I am sharing my solution and explanation for Problem 1807A – Plus or Minus. This is a very beginner-friendly problem but a good warm-up to understand simple conditions in programming.

Problem Statement

You are given three integers a, b, and c.

You need to determine whether:

a + b = c → then print "+"

a — b = c → then print "-"

It is guaranteed in the test cases that one of these will always be true.

Approach

The logic is very simple:

Read the number of test cases t.

For each test case, take three integers a, b, c.

Check:

If a + b == c → print "+"

Else if a — b == c → print "-"

Since the problem guarantees that one of them will always hold true, we don’t need to worry about any other condition.

Code (Java) import java.util.Scanner;

public class problem1807A { public static void main(String [] args){ Scanner sc = new Scanner(System.in); int t = sc.nextInt(); // number of test cases

while(t-- > 0){
        int a = sc.nextInt();
        int b = sc.nextInt();
        int c = sc.nextInt();

        if (a + b == c)
            System.out.println("+");
        else if (a - b == c)
            System.out.println("-");
    }
    sc.close();
}

}

Example Dry Run Input: 3 1 2 3 3 1 2 2 3 1

Step-by-step:

For 1 2 3: 1 + 2 = 3 → Output "+"

For 3 1 2: 3 — 1 = 2 → Output "-"

For 2 3 1: 2 — 3 = -1 (not equal) 2 + 3 = 5 (not equal either) But since problem guarantees one case will be true → actual test data won’t have this situation.

Output: + - **** Complexity

Time Complexity: O(t) (since we check each test case once)

Space Complexity: O(1) (no extra storage needed)

Conclusion

This problem was a simple conditional check. It helps beginners practice reading input, applying if-else conditions, and producing correct output format.

That’s all! Happy coding and keep solving!