Solving SPOJ SPEED: A Comprehensive Guide

Thank you for the A2A. SPOJ (Sphere Online Judge) is a renowned platform for coding enthusiasts and beginners to enhance their coding skills. SPOJ categorizes problems into various sections such as ad-hoc, classic, and others, making it easier for users to find problems based on their experience level. If you're a beginner on SPOJ, you can start by typing sort6 in the URL bar to find some beginner-friendly problems, or you can try solving a problem on your own using a text editor before submitting it on SPOJ.

One of the most frequently encountered problems on SPOJ is SPEED, which requires understanding and applying concepts such as the greatest common divisor (GCD), the absolute value, and relative speed. In this article, we will delve into how to solve the SPEED problem with a detailed explanation and useful tips.

The SPEED Problem

The problem statement for SPEED is as follows: Given the speeds of two moving entities (let's call them P1 and P2) and the distance they travel, determine the number of times they meet during their journey. To solve this problem, we need to understand the underlying algorithm and how to implement it in code.

Understanding the Problem

To solve the SPEED problem, let's break it down step by step. Consider two people, P1 and P2, traveling different distances at different speeds. The absolute distance traveled by P1 is (d1), and the absolute distance traveled by P2 is (d2). Their speeds are (a) and (b) respectively.

Define the Time to Meet: Let the time taken to meet be (t), which is the same for both entities. This means: (d1 a times t) (d2 b times t)

From these equations, we can express (t) as:

[t frac{d1}{a} frac{d2}{b}]

Since (t) is the same for both entities, we can derive the following relationship between (d1) and (d2):

[d1 d2 times frac{b}{a}]

Using the Greatest Common Divisor (GCD)

The key to solving the SPEED problem lies in using the greatest common divisor (GCD) of the speeds (a) and (b). The GCD of (a) and (b) gives us the common factor in their speeds, which helps us determine the number of times they meet.

The formula to determine the number of times they meet is:

[ text{Number of meetings} leftlfloor frac{|a - b|}{text{gcd}(|a|,|b|)} rightrfloor ]

Here's a step-by-step breakdown of the formula:

Calculate the Absolute Values: Take the absolute values of the speeds to ensure they are positive. Find the GCD: Calculate the GCD of the two speeds. Compute the Relative Speed: Compute the relative speed as the absolute difference between the two speeds, i.e., (|a - b|). Divide and Floor: Divide the relative speed by the GCD and take the floor value to get the number of meetings.

Tips for Solving the Problem

When faced with similar problems, it's important to:

Stay Focused: Keep trying different approaches and test cases to find a pattern or a solution. Use Test Cases: Test the problem with different inputs to identify patterns that might help in solving the problem. Discuss with Peers: Collaborate with colleagues or discuss the problem with others for new insights and solutions.

Here is a sample implementation in C to solve the SPEED problem:

int speedProblem(long long a, long long b, long long d1, long long d2) {
    long long gcd  gcd(abs(a), abs(b));
    long long relativeSpeed  abs(a - b);
    long long meetings  (relativeSpeed / gcd);
    return meetings;
}

In conclusion, the key to solving the SPEED problem is understanding the relationship between distance, speed, and the GCD of the speeds. By applying the aforementioned formula, you can efficiently determine the number of times two entities meet during their journey. Happy coding!