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Codeforces Problem 1878C - Vasilije in Cacak

Posted by R.A.S CODERS
Code Intuition Competitive programming

Codeforces Problem 1878C - Vasilije in Cacak

Codeforces Round 900 (Div. 3), Simplified Solution to "problem of determining whether it's possible to choose k distinct integers between 1 and n such that their sum equals x?"

problem link: Codeforces Problem 1878C - Vasilije in Cacak from Codeforces.

This blog focuses on solving the Codeforces "Vasilije in Cacak" problem, where we are asked to determine whether it’s possible to select `k` distinct integers from the range `[1, n]` such that their sum equals `x`. We’ll break the problem into simple, clear steps, explain the logic, and provide an efficient solution that works for large inputs.

Problem Statement

You are given:

  • An integer `n` representing the maximum number you can choose.
  • An integer `k` representing how many distinct integers you must select.
  • An integer `x` representing the sum you want to achieve by selecting `k` distinct integers between 1 and `n`.

You need to determine:

  • Whether it is possible to select exactly `k` distinct integers from the range `[1, n]` such that their sum equals `x`.

Key Insight: Conditions for Achievable Sums

The sum of the selected integers can range between a minimum and maximum value. Let’s compute these sums:

  • Minimum Sum (`min_sum`): This is the sum of the first `k` integers: min_sum = k * (k + 1) / 2.
  • Maximum Sum (`max_sum`): This is the sum of the last `k` integers: max_sum = k * n - (k * (k - 1)) / 2.

If `x` lies within the range `[min_sum, max_sum]`, then it is possible to achieve the sum `x`. Otherwise, it is not possible.

Step 1: Finding the Range of Achievable Sums

The minimum sum is obtained by selecting the first `k` integers from the range `[1, n]`. Using the formula for the sum of an arithmetic progression, we can calculate the minimum sum as:

min_sum = k * (k + 1) / 2

The largest sum is achieved by selecting the last k integers, i.e., [ 𝑛 , 𝑛 − 1 , . . . , 𝑛 − 𝑘 + 1 ] .Again, using the arithmetic progression formula, the sum is:

max_sum = k * n - (k * (k - 1)) / 2

If `x` is between `min_sum` and `max_sum` inclusive, then the next step is to prove that every sum between `min_sum` and `max_sum` is achievable.

Step 2: Proving Every Sum Between `min_sum` and `max_sum` is Achievable

Let's demonstrate this process with an example:

  • Let `n = 8`, `k = 3`, and we are looking for sums between `min_sum = 6` and `max_sum = 21`.

Example: Incrementing the Pointers

We start by selecting the smallest `k` integers, i.e., `[1, 2, 3]` with a sum of `6`. We will now increment the last pointer while keeping the other pointers fixed:

  • Sum of `[1, 2, 3]` = `6`
  • Sum of `[1, 2, 4]` = `7`
  • Sum of `[1, 2, 5]` = `8`
  • Sum of `[1, 2, 6]` = `9`
  • Sum of `[1, 2, 7]` = `10`
  • Sum of `[1, 2, 8]` = `11`

Once the last pointer reaches `n`, we move the second-last pointer while keeping the first pointer fixed:

  • Sum of `[1, 3, 8]` = `12`
  • Sum of `[1, 4, 8]` = `13`
  • Sum of `[1, 5, 8]` = `14`
  • Sum of `[1, 6, 8]` = `15`
  • Sum of `[1, 7, 8]` = `16`

Now, we move the first pointer while keeping the second and third pointers fixed:

  • Sum of `[2, 7, 8]` = `17`
  • Sum of `[3, 7, 8]` = `18`
  • Sum of `[4, 7, 8]` = `19`
  • Sum of `[5, 7, 8]` = `20`
  • Sum of `[6, 7, 8]` = `21`

By following this process, we can generate all the sums between `6` and `21`, proving that every sum in this range is achievable.

Algorithm Steps

  1. Start with the smallest configuration: Select the first `k` integers, i.e., `[1, 2, ..., k]`.
  2. Increment the last pointer: Move the last pointer step by step until it reaches `n`, while keeping the first `k-1` pointers fixed.
  3. Increment the second last pointer: Once the last pointer reaches `n`, increment the second last pointer and move it until it reaches `n-1`, while keeping the other pointers fixed.
  4. Continue the process: Repeat this process for each pointer, ensuring no duplicates and valid selections, until the first pointer reaches `n-k+1`.

Code Implementations

Below are the implementations in Java and C++.

JAVA


import java.util.*;
import java.math.BigInteger;

public class VasilijeInCacak {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt(); // Read the number of test cases

        while (t-- > 0) {
            BigInteger n = sc.nextBigInteger(); // Read n
            BigInteger k = sc.nextBigInteger(); // Read k
            BigInteger x = sc.nextBigInteger(); // Read x

            // Calculate the minimum and maximum possible sums
            BigInteger one = BigInteger.ONE;
            BigInteger min_sum = k.multiply(k.add(one)).divide(BigInteger.valueOf(2)); // Sum of the first k integers
			BigInteger max_sum = k.multiply(n).subtract(k.multiply(k.subtract(one)).divide(BigInteger.valueOf(2))); // Sum of the largest k integers

            // Check if x lies within the range [min_sum, max_sum]
            if (x.compareTo(min_sum) >= 0 && x.compareTo(max_sum) <= 0) {
                System.out.println("YES");
            } else {
                System.out.println("NO");
            }
        }

        sc.close();
    }
}

Conclusion

In this blog, we analyzed how to solve the "Vasilije in Cacak" problem by:

  • Finding the range of achievable sums using the arithmetic progression formula.
  • Proving that every sum between the minimum sum and the maximum sum is achievable by incrementing pointers systematically.

This approach ensures an efficient solution to the problem, making it easier to determine if a specific sum is achievable within the given constraints.

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